Perfect Square Function

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Perfect Square Function

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Source: 2021 APMO P5

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Miku3D

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Miku3D

#1 h Jun 9, 2021, 6:34 AM • 3 Y

Y by centslordm, oVlad, megarnie

Determine all Functions $f:\mathbb{Z} \to \mathbb{Z}$ such that $f(f(a)-b)+bf(2a)$ is a perfect square for all integers $a$ and $b$ .

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gghx

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gghx

#2 h Jun 9, 2021, 6:37 AM • 4 Y

Y by centslordm, tck_darkness, ZHEKSHEN, luutrongphuc

Let $P(a,b): f(f(a)-b)+bf(2a)$ is a perfect square.

Lemma: $f(0)=0$ , $f(a)$ is a perfect square

Let $f(0)=c$
$P(0,f(0)):c^2+c=k^2$ for some $k$ so $(2c+1)^2=(2k)^2+1$ .
The only time two perfect squares can be consecutive is $0,1$ so $k=0$ and thus $c=0/-1$

If $f(0)=-1$ ,
$P(0,-1-b): f(b)+b+1$ is a perfect square, $P(a,0): f(f(a))$ is a perfect square.
Thus, $f(f(a))+f(a)+1$ is a perfect square.
If $f(a)$ is $1\pmod{4}$ , then $f(f(a))+f(a)+1\equiv 2/3 \pmod{4}$ , so it cannot be a perfect square.
Thus $f(a)$ is not $1 \pmod{4}$

Now $P(0,0): f(-1)$ is a perfect square hence it is $0\pmod{4}$ .
But $P(-1,f(-1)): f(-1)f(-2)-1$ is a perfect square but it is $3$ mod 4, contradiction.

This proves that $f(0)=0$ .
Now $P(0,-a): f(a)$ is a perfect square.

Case 1: $f$ has a positive zero.
Suppose $f(c)=0, c>0$ .
Then $P(a,f(a)-c): (f(a)-c)f(2a)=0$ .
Taking $a=c$ , since $f(c)-c=-c<0$ , $f(2c)=0$ .
Similarly, $f(4c),f(8c),...$ are all $0$ .

Now $P(a,f(a)-2^kc): (f(a)-2^kc)f(2a)=0$ . Since $f(a)-2^kc$ is negative for large enough $k$ , we must have $f(2a)=0$ for all $a$ . This gives the solution $f(2x)=0$ , $f(2x+1)=g(x)^2$ for any function $g:\mathbb{Z}\rightarrow \mathbb{Z}$ .

Case 2: $f(a)>0$ for any $a>0$
$P(a,f(a)-2a): (f(a)-2a+1)f(2a)$ is a perfect square. Hence for positive $a$ , $f(a)-2a+1$ is a perfect square.
Take $k=\frac{p+1}{2}$ where $p$ is prime, since $f(k)$ and $f(k)-p$ are both perfect squares, $f(k)=k^2$ . So $f(a)=a^2$ for infinitely many $a$ .

Also, this means $f(a)\le a^2$ for positive $a$ , as bigger than that, the difference of squares would be more than $2a-1$ so $f(a)-2a+1$ won't be a perfect square.

Take any $x,y$ such that $f(x)=x^2,f(y)=y^2$ and $x,y$ positive.
$P(x,x^2-y): y^2+(x^2-y)f(2x)=y^2-yf(2x)+x^2f(2x)$ is a perfect square. Fix $x$ and increase $y$ towards infinity, we have that $4(y^2-yf(2x)+x^2f(2x))=(2y-f(2x))^2-f(2x)^2+4x^2f(2x)$ which means $f(2x)=4x^2$ .

Thus, taking any integer $a$ , $P(x,x^2-a): f(a)+(x^2-a)(4x^2)$ is a perfect square, so $4x^4-4ax^2+f(a)=(2x^2-a)^2+f(a)-a^2$ is a perfect square. Taking $x$ to infinity, we have $f(a)=a^2$ for all integers $a$ , the second solution.

1/3 done during the test, 1/3 done at home, 1/3 my senior told me

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TheUltimate123

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TheUltimate123

#3 h Jun 9, 2021, 8:15 AM • 2 Y

Y by centslordm, HamstPan38825

Say $f$ is even-vanishing if it sends all evens to zero, and let $Q$ be the set of all $q$ such that $|q|=1$ or $|q|$ is prime.

The answer is $f(x)\equiv x^2$ or $f$ is an even-vanishing function that sends odds to perfect squares. All such functions clearly work, so we show they are the only solutions. Let $P(a,b)$ denote the assertion.

From $P(0,f(0))$ , we know $f(0)+f(0)^2$ is a perfect square. If $f(0)\ne0$ , then we can bound this between $f(0)^2$ and $(f(0)+1)^2$ , so $f(0)=0$ .
Applying $P(0,-x)$ , we find that $f(x)$ is a perfect square for all $x$ .
By $P(x,f(x)-2x)$ ,we have $f(2x)(f(x)-2x+1)$ is always a perfect square. In particular, for all $x$ , either $f(2x)=0$ or $f(x)-2x+1$ is a square. Call this property $(\star)$ .

Claim 1: For $q\in Q$ , either $f(q+1)=0$ or $f(\frac{q+1}2)=(\frac{q+1}2)^2$ .

Proof. By taking $x=\frac{q+1}2$ in $(\star)$ , we know that $f(q+1)=0$ or $f(\frac{q+1}2)-q$ is a square. In the latter case, take $f(\frac{q+1}2)=u^2$ and $f(\frac{q+1}2)-q=v^2$ for $u,v\ge0$ , so $q=u^2-v^2=(u-v)(u+v)$ . It follows that $u+v=|q|$ , and the conclusion readily follows. $\blacksquare$

Claim 2: If $n\equiv1\pmod4$ , if $f(n+1)=0$ , then $f$ is even-vanishing.

Proof. Take $P(f(a),f(a)-n-1)$ for arbitrary $a$ . Then $f(2a)(f(a)-n-1)$ is a square, so either $f(2a)=0$ or $f(a)-n-1$ is a square. In the latter case, $n+1\equiv2\pmod4$ is the difference between two squares, absurd. $\blacksquare$

Claim 3: If $f(n)=0$ for any $n\ne0$ , then $f$ is even-vanishing.

Proof. Take $P(f(a),f(a)-n)$ for arbitrary $a$ . Then $f(2a)(f(a)-n)$ is a square, so either $f(2a)=0$ or $f(a)-n-1$ is a square.

Assume for contradiction $f$ is not even-vanishing, so we may choose a large prime $q\equiv1\pmod4$ with $f(q+1)\ne0$ by Claim 2, i.e.\ $f(\frac{q+1}2)=(\frac{q+1}2)^2$ by Claim 1. Select $a=\frac{q+1}2$ , so $(\frac{q+1}2)^2-n$ is a square. For $q\gg n$ this is absurd. $\blacksquare$

Assume $f$ is not even-vanishing, so the goal is to show $f(x)\equiv x^2$ .

Claim 4: $f(\frac{q+1}2)=(\frac{q+1}2)^2$ for all $q\in Q$ .

Proof. From Claim 3 above, we know $f(n)\ne0$ for $n\ne0$ , and thus the desired result follows from Claim 1. $\blacksquare$

It is now easy to generalize via induction; first, we verify the following:

Claim 5: If $f(n)=n^2$ then $f(2n)=4n^2$ .

Proof. There are infinitely many $k$ with $f(k)=k^2$ by Claim 4; for each $k$ , consider $P(n,f(n)-k)$ . Then $k^2-f(2n)k+n^2f(2n)$ is a perfect square. Since this is a perfect square for infinitely many $k$ , it must itself be a perfect square trinomial in $k$ , so $f(2n)^2=4n^2f(2n)$ , i.e.\ $f(2n)=4n^2$ . $\blacksquare$

Now we proceed by induction to show $f(n)=f(-n)=n^2$ for $n\ge0$ ; base cases $n=0,1,2,3$ readily follow from Claim 4. Let $a=n-1$ , and consider $P(f(a),f(a)-n)$ : we know $f(n)+4a^2(a^2-n)$ is a square.

Recall from $(\star)$ that $f(n)-2n+1$ is a square; hence $0\le f(n)\le n^2$ . Then it is easy to verify that $\left(2a^2-n-1\right)^2<f(n)+4a^2(a^2-n)\le\left(2a^2-n\right)^2,$ so equality holds on the right and $f(n)=n^2$ .

Analogously, $P(f(a),f(a)+n)$ will show $f(-n)=n^2$ , so the induction is complete.

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MarkBcc168

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MarkBcc168

#4 h Jun 9, 2021, 9:20 AM • 2 Y

Y by centslordm, Mango247

All answers are the benign $f(x) = x^2$ and any $f$ that sends $\text{even}\mapsto 0$ and $\text{odd}\mapsto\text{perfect square}$ , all of which clearly works. We divide the proof into multiple claims.

Claim: $f(0)=0$ and $f(x)$ is perfect square for any $x$

Proof: Plugging $a=0$ and $b=f(0)$ gives $f(0)^2+f(0)$ is a perfect square, which is enough to force that $f(0)=0,-1$ . To rule out $-1$ , we plug in $a=0$ to find out that $f(x)+x+1$ is a perfect square. Moreover, plugging in $b=0$ gives $f(f(x))$ is a perfect square. Thus, we have that
$\begin{align*} f(f(f(2))) + f(f(2)) + 1 &= \text{ perfect square} \\ f(f(2)) + f(2) + 1 &= \text{ perfect square} \\ f(2) + 3 &= \text{ perfect square} \\ \end{align*}$ Considering the first equation in mod $4$ forces $f(f(f(2))) \equiv f(f(2)) \equiv 0\pmod 4$ . The second equation then gives $f(2)\equiv -1,0\pmod 4$ , and lastly, the third equation gives the contradiction. Hence, $f(0)=0$ .

Finally, plugging in $a=0$ gives the second assertion. $\blacksquare$

Now, we give the main step.

Claim: Either one of the following assertions must be true.

There exists an infinite sequence of positive integers $x_1<x_2<\hdots$ such that $f(2x_1) = f(2x_2) = \hdots = 0$ .
There exists an infinite sequence of positive integers $x_1<x_2<\hdots$ such that $f(x_i) = x_i^2$ and $f(2x_i)\ne 0$ for all $i$ .

Proof: The main idea of the proof hinges from the substitution $b=f(a)-2a$ . This gives
$f(2a) + (f(a)-2a)f(2a) = f(2a)(f(a)-2a+1) \text{ is a perfect square}.$ Therefore, for each $a$ , either $f(2a)=0$ or $f(a)-2a+1$ is a perfect square. Motivated by this, we commit to select $a = \tfrac{p+1}{2}$ for an odd prime $p$ . Thus, either $f(p+1)=0$ or $f\left(\tfrac{p+1}{2}\right) = p$ is a perfect square.

However, if the latter happens, then both $f\left(\tfrac{p+1}{2}\right)$ and $f\left(\tfrac{p+1}{2}\right)-p$ must be a perfect square, which forces (by difference of squares) $f\left(\tfrac{p+1}{2}\right) = \left(\tfrac{p+1}{2}\right)^2$ . In conclusion, for any prime $p$ , either $f(p+1)=0$ or $f\left(\tfrac{p+1}{2}\right) = \left(\tfrac{p+1}{2}\right)^2$ , which is enough to imply the claim. $\blacksquare$

Of course, we do both cases separately. The following claim finishes the first case.

Claim: Suppose the first assertion: there exists an infinite sequence of positive integers $x_1<x_2<\hdots$ such that $f(2x_1) = f(2x_2) = \hdots = 0$ . Then, $f(\text{even}) = 0$ .

Proof: We plug in $b = f(a)-2x_i$ to get that $(f(a)-2x_i)f(2a)$ is a perfect square. However, if $2x_i > f(a)$ , then it forces $f(2a)=0$ for sign reason. $\blacksquare$

Now, assume the second case: there exists an infinite sequence of positive integers $x_1<x_2<\hdots$ such that $f(x_i) = x_i^2$ and $f(2x_i)\ne 0$ for all $i$ . We need the following well-known lemma.

Lemma: Suppose that $a,b$ are integers such that $a^2\ne 4b$ . Then, there are finitely many $x$ such that $x^2+ax+b$ is perfect square.

Proof: Just complete the square. I don't bother writing the details. $\blacksquare$

To finish, we prove the following two claims.

Claim: For any $i$ , $f(2x_i) = 4x_i^2$ .

Proof: Vary $j$ through all positive integers and plug in $a=x_i$ and $b=x_i^2-x_j$ to obtain that
$f(x_j) + (x_i^2-x_j)f(2x_i) = x_j^2 - x_jf(2x_i) + x_i^2f(2x_i) \text{ is a perfect square},$ so by the lemma, $f(2x_i)^2 = 4x_i^2f(2x_i)\implies f(2x_i) = 4x_i^2$ (reminder: $f(2x_i)\ne 0$ ). $\blacksquare$

Claim: $f(k)=k^2$ for any $k$

Proof: Vary $i$ through all positive integers and plug in $a = x_i$ and $b = x_i^2-k$ to obtain that
$f(k) + 4x_i^2(x_i^2-k) = (2x_i^2)^2 + (2x_i^2)\cdot 2k + f(k) \text{ is perfect square},$ so by the lemma once again, we get that $(2k)^2 = 4f(k)\implies f(k)=k^2$ . $\blacksquare$

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Mr.C

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Mr.C

#5 h Jun 9, 2021, 11:45 AM • 2 Y

Y by centslordm, Wizard0001

My solution at the exam :

Let
$P(a,b)^2=f(b)+f(a)f(2a)-bf(2a)$ from $P(0,0)^2$ we have $f(0) \in \{0,-1\}$ .
Assume $f(0)=0$ . now if $a>0,f(a)=0$ we have from $P(a,a)^2$ that $-af(2a)$ is a perfect square.
Also from $P(0,2a)^2$ we have $f(2a)$ is a perfect square aswell.
Hence $f(2a)=0$ . now from $P(x,2^n a)^2$ we have $f(x)f(2x)-2^naf(2x)$ is a perfect square. hence $f(2x)=0$ for all integers $x$ .
now the condition goes : $P(a,b)^2=f(b)$ hence $f(2k+1)=g(k)^2,f(2k)=0$ works for every arbitrary function $g$ .
Now if $f(a)$ is non zero for positive $a$ we have :
$P(a,2a)^2=f(2a)(1+f(a)-2a)$ $P(0,2a)^2=f(2a)>0$ hence $T(a)^2=1+f(a)-2a$ also let $f(a)=g(a)^2$ clearly $(g(a)-1)^2 \ge f(a)-2a+1$ hence $a \ge g(a)$ . which gives $f(1)=1$ .
Now from $P(a,1)=1+f(a)f(2a)-f(2a)$ is a perfect square. if $f(2a)=1$ then $f(2a)-4a+1 <-1$ so it is not a perfect square. hence :
$(g(a)g(2a)-1)^2 \ge f(a)f(2a)-f(2a)+1$ hence :
$f(2a) \ge 4f(a)$ which gives clearly :
$f(2^n) \ge 4^n$ but we also had $g(a) \le a$ so $f(2^n)=4^n$ . now from $P(2^n,b)^2=f(b)+(2^{2n+1})^2-2b.2^{2n+1}$ so we have :
$P(2^n,b)^2+b^2=f(b)+(2^{2n+1} -b)^2$ so $f(b)-b^2$ is the difference of infinitely many perfect squares hence it is $0$ . so $f(x)=x^2$ is also a solution.
Now if $f(0)=-1$ we have :
$P(a,0)^2=f(a)f(2a)-1$ hence if $q$ is a positive devisor of $f(a)$ for some $a$ , then $q=4k+2$ or $q=4k+1$ .
Now from $P(0,a)^2=f(b)+b+1$ now $f(4k)+1+4k$ is a perfect square. so $f(4k) \in \{4a,4a-1\}$ which is imposible if $f(4k)>0$ so $f(4k)$ should be negative.
Now we prove $f(2a) \le 0$ which is trivial since $f(4a)f(2a)-1$ is perfect square hence non negative. assume $f(2a)=0$ then we have $f(b)$ is a perfect square itself so $f(4k)=0$ hence $f(0)=0$ and it is a contradiction. so $f(2a)<0$ now since
$f(2a)f(a)-1$ is a perfect square we get that for all integers $n$ we have $f(n) \le 0$ now we have $f(1)+2$ is a perfect square.
So $f(1) \in \{-1,-2\}$ if $f(1)=-2$ we have $-2f(2)-1$ is a perfect square. also $f(2)+3$ is a perfect square aswell. which is impossible .
Hence $f(1)=-1$ and it also gives $f(2)=-2$ . now from $P(1,1)^2=-1+4$ is a perfect square which is a contradiction.
so in this case we have no solutions.

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Orestis_Lignos

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Orestis_Lignos

#6 h Jun 9, 2021, 11:51 AM • 2 Y

Y by centslordm, SerdarBozdag

Let $P(x,y)$ be the assertion that $f(f(x)-y)+yf(2x)$ is a perfect square for all $x,y \in \mathbb{Z}$ .

Then, we proceed with some Claims.

Claim 1: $f(0)=0$ .
Proof: Considering $P(0,f(0))$ we obtain that $f^2(0)+f(0)$ is a perfect square, hence $f(0) \in \{-1,0 \}$ .

Suppose FTSOC that $f(0)=-1$ . Then, $P(0,-x-1)$ implies $f(x)+x+1$ is a perfect square. In addition, by $P(x,f(x)$ we obtain that $f(x)f(2x)-1$ is also a perfect square.

Now, take a $x \equiv 0 \pmod 4$ . Then, $f(x)+x+1 \equiv f(x)+1 \pmod 4$ , therefore $f(x) \equiv 0 \, \, \rm or \, \, 3 \pmod 4$ .

If $f(x) \equiv 0 \pmod 4$ , then $f(x)f(2x)-1 \equiv 3 \pmod 4$ , which can never be a perfect square.
If $f(x) \equiv 3 \pmod 4$ , then if $f(x)f(2x)-1=m^2$ , we obtain that $f(x) \mid (m^2+1)$ , hence if $p \equiv 3 \pmod 4$ is a prime dividing $f(x)$ , we obtain $p \mid (m^2+1)$ . By Fermat's Christmas Theorem, this gives $p \mid 1$ , a contradiction $\blacksquare$

Claim 2: $f(x)$ is a perfect square for all $x$ .
Proof: Consider $P(0,-x)$ $\blacksquare$

Now, we may let $f(x)=g^2(x)$ , with $g: \mathbb{Z} \rightarrow \mathbb{N}$ i.e., $g$ takes only nonnegative values.

$P(x,y)$ rewrites as $g^2(g^2(x)-y)+yg^2(2x)$ is a perfect square. Taking $y \rightarrow g^2(x)-y$ and then swapping $x$ and $y$ we obtain that $g^2(x)+(g^2(y)-x)g^2(2x)$ is a perfect square, for all $x,y \in \mathbb{Z}$ . Let this be $Q(x,y)$ .

We distinguish two cases.

Case 1: There exists a $k \neq 0$ such that $g(k)=0$ . Then, we make the following Claim:

Claim 3: $g(2k)=0$ .
Proof: Suppose not. Then, $g(2k) \neq 0$ . If there exists an $\ell >0$ such that $g(\ell)=0$ , then $Q(\ell,k)$ implies $-\ell g^2(2k)$ is a perfect square. Since $g(2k) \neq 0$ and $\ell>0$ , this expression is $<0$ , a contradiction.

Therefore, $g(x) >0$ , for all $x>0$ .

Now, take a $x>0$ and consider $Q(k,x)$ . This gives that $(g^2(x)-k)g^2(2x)$ is a perfect square. Since $g(2x)>0$ from the above discussion, we obtain that $g^2(x)-k$ is also a perfect square, which implies that $g(x)$ takes finitely many values for all $x>0$ (specifically, $g(x)$ is of the form $\dfrac{d_i+d_j}{2}$ where $d_i,d_j$ are divisors of $k$ such that $d_id_j=k$ ).

Let the maximum value of $g(x)$ for $x>0$ be $S$ . Then, fixing $y$ and taking a pretty large $x$ , we obtain that $g^2(x)+(g^2(y)-x)g^2(2y) \leq S^2+g^2(y)g^2(2y)-xg^2(2y)<0,$ a contradiction.

Hence, the Claim is proved $\blacksquare$

To the problem, the Claim implies that $g(k)=0 \Rightarrow g(2k)=0 \Rightarrow g(4k)=0 \Rightarrow \ldots g(2^tk)=0$ , for all $t>0$ .

Hence, $Q(2^tk,x)$ implies that $(g^2(x)-2^tk)g(2x)$ is a perfect square.

If, for some $x$ , $g(2x) \neq 0$ , then $g^2(x)>2^tk$ for all $t$ , which is an immediate contradiction.

Therefore, $g(2x)=0$ for all $x \in \mathbb{Z}$ . It is trivial to verify now that any function of the form $g(x) = \begin{cases} 0 & \text{if} \,\, x \,\, \text{is even} \\ \text{anything} & \text{if} \,\, x \,\, \text{is odd} \end{cases}$ satisfies.

Equivalently, we obtain that $f(x) = \begin{cases} 0 & \text{if} \,\, x \,\, \text{is even} \\ \text{any perfect square} & \text{if} \,\, x \,\, \text{is odd} \end{cases}$ is a solution to the problem.

Case 2: $g(x)=0$ only if $x=0$ . Then $g(x) >0$ when $x \neq 0$ . Considering $Q(2x,x)$ when $x \neq 0$ , we obtain that $(g^2(x)-2x+1)g^2(2x)$ is a perfect square, hence $g^2(x)-2x+1$ is also a perfect square $(*)$ .

Now, we prove a series of Claims.

Claim 4: $g(\dfrac{p+1}{2})=\dfrac{p+1}{2}$ for all odd primes $p$ .
Proof: Taking $x=\dfrac{p+1}{2}$ in $(*)$ we obtain that $g^2(\dfrac{p+1}{2})-p=a^2$ , hence $(g(\dfrac{p+1}{2})-a)(g(\dfrac{p+1}{2})+a)=p,$ implying that $g(\dfrac{p+1}{2})-a=1, \,\, g(\dfrac{p+1}{2})+a=p,$ which readily gives the desired $\blacksquare$

Claim 5: $g(2x)$ is even for all $x$ .
Proof: By $(*)$ , $g^2(2x)-4x+1$ is a perfect square. If $g(2x)$ is odd, then $g^2(2x)-4x+1 \equiv 2 \pmod 4$ , a contradiction $\blacksquare$

Claim 6: $g(2x)=2g(x)$ .
Proof: Fix a $x$ and let $g(x)=M$ and $g(2x)=N$ . Then, by considering $Q(\dfrac{p+1}{2},x)$ , we obtain that $M^2N^2-\dfrac{p+1}{2}N^2+(\dfrac{p+1}{2})^2$ is a perfect square for all odd primes $p$ .

The latter rewrites as $(\dfrac{N^2}{2}-\dfrac{p+1}{2})^2+M^2N^2-\dfrac{N^4}{4},$ where the expressions in the brackets and the one outside of them are integers due to Claim 5.

If $M^2N^2-\dfrac{N^4}{4} \neq 0$ , then by letting $p \rightarrow +\infty$ we obtain a clear contradiction, since then $\dfrac{N^2}{2}-\dfrac{p+1}{2}$ should attain finitely many values, a contradiction.

Therefore, $M^2N^2-\dfrac{N^4}{4}=0$ , which easily rearranges to $N=2M$ , i.e. $g(2x)=2g(x)$ , as desired $\blacksquare$

Claim 7: $g^2(x)=x^2$ .
Proof: Using Claim 6, $Q(x,y)$ rewrites as $g^2(x)+4(g^2(y)-x)g^2(y)$ being a perfect square.

Taking in the latter $y=\dfrac{p+1}{2}$ and fixing $x$ and rearranging, we obtain that $(\dfrac{p+1}{2}-x)^2+g^2(x)-x^2$ is a perfect square, which when $g^2(x)-x^2 \neq 0$ is a contradiction after taking $p \rightarrow +\infty$ $\blacksquare$

Therefore, we conclude that $g^2(x)=x^2$ , i.e. $f(x)=x^2$ which is a solution.

To conclude, $f(x)=x^2$ and $f(x) = \begin{cases} 0 & \text{if} \,\, x \,\, \text{is even} \\ \text{any perfect square} & \text{if} \,\, x \,\, \text{is odd} \end{cases}$ are the only solutions to the problem.

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hyay

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hyay

#7 h Jun 9, 2021, 11:55 AM • 2 Y

Y by centslordm, somebodyyouusedtoknow

The only such functions are $f(x) = x^2$ for all $x$ and $f(x) = \begin{cases} 0 & \text{if } 2 \mid x \\ g(x)^2 & \text{otherwise} \end{cases}$ for all $x$ , where $g(x)$ is any function from the integers to itself.

Let $P(a, b) = f(f(a) - b) + bf(2a)$ .

First of all, $P(0, 0) = f(0)(f(0) + 1)$ is a square. If $f(0) \neq 0, -1$ , then both $|f(0)|$ and $|f(0) + 1|$ are squares since they are relatively prime, but this is only possible when $f(0) = 0/-1$ , contradiction. If $f(0) = -1$ , then $P(0, -x - 1) = f(x) + x + 1$ and $P(x, 0) = f(f(x))$ are both squares for all $x$ . Recall that $n^2 \equiv 0/1 \pmod{4}$ for all $n$ , so since $f(2) + 3$ is a square, then $f(2) \equiv 1/2 \pmod{4}$ . Also, since $f(f(f(2))) + f(f(2)) + 1$ is a square and all of the terms are squares, then we must have $f(f(f(2))) \equiv f(f(2)) \equiv 0 \pmod{4}$ . But then $f(f(2)) + f(2) + 1 \equiv 2/3 \pmod{4}$ is a square, contradiction. So, $f(0) = 0$ .

Then, $P(0, -x) = f(x)$ is a square for all $x$ . Furthermore, $P(x, f(x) - 2x) = f(2x)(f(x) - 2x + 1)$ , so either $f(2x) = 0$ or $f(x) - 2x + 1$ is a square.

Case 1. $f(2x) = 0$ for infinitely many positive integers $x$ .

We claim that $f(2x) = 0$ for all $x$ . Assume that there exists $t$ such that $f(2t) \neq 0$ . For all $k$ with $f(2k) = 0$ , $P(t, f(t) - 2k) = (f(t) - 2k)f(2t)$ is a square. However by taking $k$ sufficiently large the value becomes negative (recall that $f(2t)$ is a square), contradiction. Hence, $f(x) = 0$ for all even $x$ , while $f(x)$ is a square for all $x$ . This is exactly the second solution.

Case 2. There exists $N$ such that $f(x) - 2x + 1$ is a square for all $x \geq N$

Suppose $x \geq N$ and $2x - 1$ is a prime, then let $f(x) = a^2$ and $f(x) - 2x + 1 = b^2$ where $a, b \geq 0$ . Then $2x - 1 = (a + b)(a - b)$ , which implies $a + b = 2x - 1$ and $a - b = 1$ , thus $a = x$ and $f(x) = x^2$ . This implies there exists infinitely many $x \geq 0$ such that $f(x) = x^2$ .

Then, for all $a$ and $x \geq 0$ with $f(x) = x^2$ , $P(a, f(a) - x) = x^2 - xf(2a) + f(a)f(2a)$ is a square. If $f(2a) = 2k + 1$ for some $k \in \mathbb{Z}$ , then notice that $(x - k - 1)^2 < P(a, f(a) - x) < (x - k)^2$ for all $x$ sufficiently large, contradiction. So, $f(2a) = 2k$ and $(x - k - 1)^2 < P(a, f(a) - x) < (x - k + 1)^2$ for all large $x$ , which means $x^2 - xf(2a) + f(a)f(2a) = (x - f(2a)/2)^2$ for all large $x$ . Comparing coefficients, we get $f(2a) = 4f(a)$ .

Now, for all $a$ with $f(a) = a^2$ , $P(a, f(a) - x) = f(x) - 4a^2x + 4a^4$ is a square. Then for all $a$ sufficiently large, we have $(2a^2 - x - 1)^2 < P(a, f(a) - x) < (2a^2 - x + 1)^2$ , so $P(a, f(a) - x) = (2a^2 - x)^2$ for all large $a$ , and thus $f(x) = x^2$ for all $x$ .

Done.

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Supercali

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Supercali

#8 h Jun 9, 2021, 1:52 PM • 3 Y

Y by centslordm, PRMOisTheHardestExam, Mango247

Kinda annoying problem ngl.

By substituting $x=f(a)-b$ , rewrite the original condition as $f(x)-xf(2a)+f(a)f(2a) \text{ is a perfect square}$ Let $P(x,a)$ denote the above. Also we will use "PS" for perfect square.

Claim 1: $f(0)=0$
Proof: $P(0,0)$ $\implies$ $f(0)(f(0)+1)$ is a PS $\implies$ $f(0) =0$ or $-1$ . Assume FTSOC that $f(0)=-1$ . Then $P(-4,0)$ $\implies$ $f(-4)-3$ is a PS; in particular, $f(-4)>0$ . Taking this modulo $4$ , we get $f(-4) \equiv 3 \text{ or } 0 \pmod 4$ . Now $P(0,-4)$ gives $f(-4)f(-8) =k^2+1$ for some $k \in \mathbb{Z}$ . It is well known that all positive factors of a number of the form $k^2+1$ are congruent to $1$ or $2$ modulo $4$ , which contradicts the previous congruence. So $f(0)=0$ . $\blacksquare$

Claim 2: $f(x)$ is a PS for all integers $x$ .
Proof: $P(x,0)$ . $\blacksquare$

Let $y=\frac{p+1}{2}$ for any odd prime $p$ . Then $P(2y,y)$ gives $f(2y)(f(y)-p)$ is a PS. This means either $f(2y)=0$ , or $f(y)-p$ is a PS. But it is easy to see that $p$ can be written in a unique way as a difference of two squares - $\left ( \frac{p+1}{2} \right)^2-\left ( \frac{p-1}{2} \right)^2$ , so we must have $f(y)=\left ( \frac{p+1}{2} \right)^2=y^2$ in this case. Using the deep, highly non-trivial, and recently discovered result known as "the infinitude of primes", we can say that either $f(2y)=0$ , or $f(2y) \neq 0$ and $f(y)=y^2$ for infinitely many positive integers $y$ .

Case I: $f(2y)=0$ for infinitely many $y>0$

Fix any arbitrary integer $a$ , and choose a $y>f(a)+200000$ satisfying $f(2y)=0$ . Then $P(2y,a)$ gives $f(2a)(f(a)-2y)$ is a PS. Since $f(a)-2y<0$ , this can only be true if $f(2a)=0$ . Thus $f$ is $0$ on even numbers, and using Claim 2, we get the following solution: $\boxed{f(x) = \left( \frac{1-(-1)^x}{2} \right ) g(x)^2 \ \ \forall x \in \mathbb{Z} }$ for any arbitrary $g: \mathbb{Z} \rightarrow \mathbb{Z}$ .

Case II $f(2y) \neq 0$ and $f(y)=y^2$ for infinitely many $y>0$

Call all such $y$ good.

Claim 3: $f(2y)=4y^2$ for all good $y$ .
Proof: Fix any arbitrary good $y$ , and let $q(t)$ be the polynomial $t^2-tf(2y)+y^2f(2y)$ . There exists a positive integer $M$ satisfying $(t-M)^2<q(t)<(t+M)^2$ for all $t>0$ (for instance, you can take $M=(y^2f(2y))^{1000}+1000+y^{999}+99f(2y)^9$ ). If $t$ is good, then by $P(t,y)$ , $q(t)$ is a PS $\implies$ $q(t)=(t+n)^2$ for some integer $n \in [-M,M]$ . But since there are infinitely many good $t$ , by another deep result known as "pigeonhole principle", there exists an integer $k \in [-M,M]$ such that $q(t)=(t+k)^2$ for infinitely many good $t$ . Since this is a polynomial equation with infinitely many roots, we can compare coefficients to get $f(2y)^2=4y^2f(2y)$ $\implies$ $f(2y)=4y^2$ since $f(2y) \neq 0$ . $\blacksquare$

Now, let $x \in \mathbb{Z}$ be arbitrary. Then for any good $y$ , $P(x,y)$ gives $f(x)-4xy^2+4y^4=f(x)-x^2+(2y^2-x)^2 \text{ is a PS}$ Assume $f(x)-x^2 = m \neq 0$ for some $x$ . Take a large good $y$ , then there must be some PS at a distance of $m$ from $(2y^2-x)^2$ . Since the nearest perfect square to $(2y^2-x)^2$ is $2(2y^2-x)-1$ away, we must have $m \geq 2(2y^2-x)-1$ , which clearly cannot hold for sufficiently large $y$ , contradiction! Therefore we get the solution $\boxed{f(x)=x^2 \ \ \forall x \in \mathbb{Z}}$ $\blacksquare$

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MathForesterCycle1

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MathForesterCycle1

#9 h Jun 9, 2021, 2:08 PM • 1 Y

Y by centslordm

dame dame

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jasperE3

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jasperE3

#11 h Jun 15, 2021, 1:39 AM

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Let $\mathbb S=\{x^2\mid x\in\mathbb Z\}$ , and let $P(a,b)$ be the assertion $f(f(a)-b)+bf(2a)\in\mathbb S$ .
$P(0,f(0))\Rightarrow f(0)+f(0)^2=k_1^2$
If $f(0)\ge1$ then $f(0)^2<k_1^2<(f(0)+1)^2$ , contradiction. Similarly, if $f(0)\le-2$ then $(f(0)+1)^2<k_1^2<f(0)^2$ , another contradiction, so this leaves $f(0)\in\{-1,0\}$ .

$\textbf{Case 1: }f(0)=-1$
$P(0,-1-f(x))\Rightarrow f(f(x))+f(x)+1\in\mathbb S$ , so $f(x)+1$ is a difference of squares since $P(x,0)\Rightarrow f(f(x))\in\mathbb S$ . Since squares are always either $0$ or $1\pmod4$ , we have that $f(x)+1\not\equiv2\pmod4$ , so $f(x)\not\equiv1\pmod4$ . Note that if $f(x)\in\mathbb S$ now, we must have $f(x)\equiv0\pmod4$ . An example is $P(0,0)\Rightarrow f(-1)\equiv0\pmod4$ . However, then we have $P(-1,f(-1))\Rightarrow f(-1)f(-2)-1=k_2^2$ , so taking this equation $\pmod4$ gives us $k_2^2\equiv3$ , absurd.

$\textbf{Case 2: }f(0)=0$
$P(0,-x)\Rightarrow f(x)\in\mathbb S\Rightarrow f(x)\ge0$

$\textbf{Case 2.1: }\exists k>0:f(k)=0$
$P(k,-2k)\Rightarrow(1-2k)f(2k)\ge0$ , so $f(2k)\le0$ , hence $f(2k)=0$ with $2k>0$ . Simple induction gives $f(2^nk)=0$ for all $n\in\mathbb N$ .
$P(x,f(x)-2^nk)\Rightarrow f(2x)(f(x)-2^nk)\ge0$ , but since $f(2x)\ge0$ and $f(x)-2^nk$ is negative for sufficiently large $n$ , we must have $f(2x)=0$ . Combined with the fact that $f$ is always square, we have the solution $\boxed{f(x)=\left(\frac{1-(-1)^x}2\right)h(x)}$ , where $h$ is any function from $2\mathbb Z+1\to\mathbb S$ , which works since $P(a,b)$ reduces to $f(f(a)-b)\in\mathbb S$ , which is true since $f$ is always a square.

$\textbf{Case 2.2: }f(x)>0\forall x>0$
$P(x,f(x)-2x)\Rightarrow f(2x)(f(x)-2x+1)\in\mathbb S$
Choose $x=\frac{p+1}2$ for an odd prime $p$ , then $f(p+1)(f(x)-p)\in\mathbb S$ , and since $f(p+1)$ is square, either $f(p+1)=0$ or $f(x)-p$ is a square. The latter must be true in this case. Since $f(x)$ and $f(x)-p$ are both square, we get $p=m^2-n^2$ for $m,n$ such that $f(x)=m^2$ and $f(x)-p=n^2$ . Then, since $p$ is prime, we require $m=n+1$ , so $f(x)=n^2+2n+1=n^2+p$ , so $p=2n+1=2x-1$ , hence $x=n+1=m$ . Then $f(x)=m^2=x^2$ .
To summarize, $\exists g:\mathbb N\to\mathbb N$ strictly increasing such that $f(g(n))=g(n)^2$ for all $n\in\mathbb N$ (where $g(n)=\frac{p+1}2$ and $p$ is the $n$ th odd prime).

Now $P(x,f(x)-2x)\Rightarrow f(x)-2x+1\in\mathbb S\Rightarrow f(x)\le x^2$
$P(g(x),g(x)^2-g(y))\Rightarrow g(y)^2-g(y)f(2g(x))+g(x)^2f(2g(x))\in\mathbb S$ , setting $y$ such that $g(y)$ is sufficiently large gives that $f(2x)=4x^2$ .
Then $P(x,y)$ becomes $Q(x,y):f(f(x)-y)+4x^2y$ .
$Q(x,f(x)-y)\Rightarrow f(y)+4x^2f(x)-4x^2y\in\mathbb S$ , setting $x=g(n)$ large enough gives us that $(2g(n)^2-y)^2+f(y)-y^2\in\mathbb S$ , so $\boxed{f(x)=x^2}$ , which also works since $f(f(a)-b)+bf(2a)=(a^2+b)^2\in\mathbb S$ .

This post has been edited 4 times. Last edited by jasperE3, Apr 18, 2025, 6:59 AM

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ZETA_in_olympiad

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ZETA_in_olympiad

#12 h Jun 3, 2022, 4:19 PM • 2 Y

Y by Mango247, Mango247

Let $P(a,b)$ denote the given assertion and let $\mathcal{T}=\{t^2: t\in \mathbb{Z}\},$ i.e \ the set of squares.

$P(0,f(0))$ gives $f(0)+f(0)^2\in \mathcal{T}.$ So $f(0)=0$ or $f(0)=-1.$ Assume the latter, $P(0,a)$ gives $f(a)+a+1 \in \mathcal{T}.$ And $P(a,0)$ gives $f(f(x))\in \mathcal{T}.$ But comparing $P(0,2)$ then $P(0,f(f(2)))$ then $P(0,f(2))$ gives $f(2)\equiv 1\pmod{4}$ and $f(2)\equiv -1$ or $0 \pmod{4},$ absurd. So $f(0)=0.$ And $P(0,-a)$ gives $f(a)\in \mathcal{T}.$

Incomplete...

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ZETA_in_olympiad

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ZETA_in_olympiad

#13 h Jun 4, 2022, 11:01 AM

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Completing...

A. $f(x)=0$ for some $x>0:$ Fix some arbitrary $u$ and let $f(v)=0.$ $P(u,f(u)-v)$ gives $f(2u)(f(u)-v)\in \mathcal{T}.$ Taking large $v$ forces $f(2u)=0.$ So $f(x_{2\mid x})=0$ and $f(x_{2\nmid x})\in \mathcal{T}$ which works.

B. $f(x)> 0~~\forall x>0:$ Then let $p$ be a prime. $P(\frac{p+1}{2}, f(\frac{p+1}{2})-p+1)$ yields $(f(\frac{p+1}{2})-p)(f(p+1))\in \mathcal{T}.$ Taking $p\to \infty$ gives $f(p+1)\in \mathcal{T}.$ It follows that $f(\frac{p+1}{2})-p \in \mathcal{T}$ and $f(\frac{p+1}{2})\in \mathcal{T}.$ Let $f(\frac{p+1}{2})=a^2$ and $f(\frac{p+1}{2})-p=b^2.$ Then solving for $p$ gives $a=\frac{p+1}{2}.$ So $f(\frac{p+1}{2})=(\frac{p+1}{2})^2.$

Take fixed $x$ and an arbitrary $y$ such that $f(x)=x^2$ and $f(y)=y^2.$ $P(x,f(x)-y)$ gives $y^2+x^2f(2x)-yf(2x).$ Note that taking $y\to \infty$ (it is not hard) forces $f(2x)=4x^2.$ Now $P(y,f(y)-x)$ gives $f(x)-x^2+(2y^2-x)^2\in \mathcal{T}.$ Again $y\to \infty$ gives $f(x)=x^2,$ which works.

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Inconsistent

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Inconsistent

#16 h Mar 5, 2023, 6:38 PM

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Solution is $f(x) = 0$ for all even $x$ and equals any perfect square for all odd $x$ .

First, notice $f(0)^2 + f(0)$ is a perfect square. So if $x = f(0), y^2 = f(0)^2+f(0)$ , then $(2x+1)^2-(2y)^2 = 1$ so $y = 0$ so $x = 0, -1$ .

If $x = -1$ , then plugging $b = f(a)$ gives $f(a)f(2a)-1$ is a perfect square. Plugging $a = 0$ gives $f(x)+x+1$ is a perfect square. Thus $f(4k) \equiv 0, 3 \pmod 4$ , but if $4 \mid f(4k)$ then $f(4k)f(8k) - 1 \equiv 3 \pmod 4$ contradiction, so $f(4k) \equiv 3 \pmod 4$ for all $k$ . However, $f(4k+2) \equiv 1, 2 \pmod 4$ for all $k$ but if $f(4k + 2) \equiv 1 \pmod 4$ then $f(4k+2)f(8k+4) - 1 \equiv 2 \pmod 4$ contradiction so $f(4k+2) \equiv 2 \pmod 4$ for all $k$ . However since with $b = 0$ we have $f(f(a))$ is a perfect square for all $a$ , it follows that $f(f(4k+2))$ is a perfect square but is $2 \pmod 4$ , which is a contradiction.

Thus $x = 0$ . Assume there exists $c \neq 0$ such that $f(c) = 0$ . With $a = 0$ we get $f(x)$ is a perfect square for all $x$ . Now notice plugging $b = f(a) - 2a$ gives $f(2a)(f(a)-2a+1)$ is a perfect square for all $a$ . So either $f(2a) = 0$ or $f(a) \geq 2a - 1$ . Consider the set $S$ of $a$ such that $f(2a) \neq 0$ . If $S$ is infinite then it is unbounded. However plugging in, for $c \neq 0$ such that $f(c) = 0$ , $a = f(a) - c$ gives $(f(a)-x)f(2a)$ is a perfect square, so $f(a) - x$ is a perfect square for all $a$ in $S$ . However since $f(a)$ and $f(a) - x$ are both perfect squares for fixed $x$ , $f(a)$ is bounded. However since $f(a) \geq 2a - 1$ for all $a \in S$ , it follows that $S$ is bounded, contradiction. Thus $S$ is finite, so for all $a > N$ for some fixed $N$ , we have $f(2a) = 0$ .

Now suppose $f(2a) \neq 0$ for some $a$ . Then plugging in $c = -b$ gives $f(f(a)+c) - cf(2a) \geq 0$ . However for large positive $c$ of certain parity, $f(f(a)+c) = 0$ , so it follows that $f(2a) = 0$ . Contradiction. Thus $f(2a)$ is $0$ for all $a$ , giving the desired solution.

Finally consider the case where $f(x) \neq 0$ for all $x \neq 0$ . Then as before, it follows that $f(a)-2a+1$ is a perfect square for all $a$ . So $f(1) = 1, f(2) = 4$ and $f(a) \leq a^2$ . Thus $f(x) - 4x+4$ is a perfect square with $a = 1$ . It follows for $n = 3$ that $f(3) = 9$ , $f(4) = 16$ by size. Thus again we have $f(x) - 16x + 64$ is a perfect square for all $x$ . Again by size and $f(a) - 2a + 1$ , $f(x) = x^2$ for all $x \leq 8$ . It is easy to verify from small cases that $f(9) = 81$ and $f(10) = 100$ . Now we strong induct: supposing it is true up to $2a$ , we have

$f(c) - 4a^2c+4a^4$ is a perfect square, but LHS differs at most $c^2$ from $(2a^2 - c)^2$ . Thus for $c = 2a+1$ , we have that $f(c) = c^2$ , $4a^2+4a+1 \geq 2(2a^2 - 2a - 1) - 1$ (possibly true), or $4a^2+4a+1 \geq 4(2a^2-2a-1)-4 = 8a^2 - 8a-8 \Longleftrightarrow 4a^2-12a-9 \leq 0$ which fails for $a \geq 5$ . But in the middle case, it follows that $f(2a+1)^2 = 8a+4$ , but then $f(2a+1)-16(2a+1)+64 < 0$ contradiction. Thus $f(2a+1) = (2a+1)^2$ . Now for $c = 2a+2$ , we have that $f(c) = c^2$ , $4a^2+8a+4 \geq 2(2a^2-2a-1)-1$ (possibly true), or $4a^2+8a+4 \geq 4(2a^2-2a-1)-4 = 8a^2-8a-8 \Longleftrightarrow 4a^2-16a-12 \leq 0$ which fails for $a \geq 5$ . But in the middle case, it follows that $f(2a+2)^2 = 12a + 7$ . This is impossible mod $4$ , so it must be that $f(2a+2) = (2a+2)^2$ . Now the inductive step is complete, so $f(x) = x^2$ for all positive $x$ .

Plugging $b = f(a)+c$ for positive $a$ gives $f(-c) + (f(a)+c)f(2a) = f(-c)+ 4a^2c+4a^2$ is a perfect square, so through the analogous argument as above (same rule used), the same conclusion follows through induction.

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CircleInvert

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CircleInvert

#19 h Dec 4, 2023, 11:42 PM

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There are two types of solutions: one is $f(x)=x^2$ and the other is $f(x)=0$ for all even $x$ and for all odd $x$ we choose $f(x)$ to be any perfect square (choice may be dependent on $x$ ). Clearly both types of solutions work. It remains to prove they are the only ones.

Plug in $a=0$ and $b=f(0)$ . This gives $f(0)+(f(0))^2=f(0)(f(0)+1)$ is a perfect square. Now, observe that the factors are relatively prime; thus either one is $0$ or both are positive perfect squares, but no two positive perfect squares differ by exactly $1$ , so either $f(0)=0$ or $f(0)=-1$ .

Let's deal with $f(0)=-1$ : plugging in $b=f(a)$ , we get $f(a)f(2a)-1$ is always a perfect square, say $f(a)f(2a)=k^2+1$ . Then $f(a)$ must be either $1$ or $2$ mod $4$ for all $a$ , but now if we plug in $a=0$ and $b=3$ , we get $f(-4)-3$ is a perfect square, so it is $0$ or $1$ mod $4$ , but this gives that $f(-4)$ is $0$ or $3$ mod $4$ , contradiction.

Hence $f(0)=0$ .

Plugging in $a=0$ we get that the range of $f$ is a subset of the set of perfect squares.

Assume that there exists $2c$ such that $f(2c)\ne 0$ (otherwise we are in a solution set we have already identified). Then as $f(c)$ is a perfect square, $f(c)>0$ .

Plugging in $a=c$ , we get $f(f(c)-b)+bf(2c)$ is always a perfect square, and thus $f(f(c)-b)+bf(2c)\ge 0$ . Then $f(f(c)-b)\ge -bf(2c)$ . Hence, if we send $b$ to $-\infty$ , this shows that as $x\to\infty$ we have $f(x)\to\infty$ .

Suppose for the sake of contradiction that $k\ne 0$ and $f(k)=0$ . Then then, letting $b=f(a)-k$ we get that for every $a$ we have $(f(a)-k)f(2a)$ is a perfect square. For sufficiently large (positive) $a$ , we have $f(2a)>0$ , and of course $f(2a)$ is a perfect square, so this implies that $f(a)-k$ is a perfect square. However, as $a\to\infty$ , $f(a)\to\infty$ , and $f(a)$ is also a perfect square, so this would imply arbitrarily large perfect squares that are a constant $k$ apart, which is absurd, hence we get our contradiction, and conclude that $f$ is positive besides at $0$ .

Now, letting $b=f(a)-2a$ , we get that $f(2a)(1+f(a)-2a)$ is a perfect square. This means that for $a\ne 0$ , $1+f(a)-2a$ is a perfect square (and actually $a=0$ can be checked seperately so this statement just holds for all $a$ ). Now, suppose $2a-1$ . Then let $1+f(a)-2a=A^2$ and $f(a)=B^2$ where $A,B\ge 0$ . We then have $2a-1=B^2-A^2=(B-A)(B+A)$ , so $B-A=1$ and $B+A=2a-1$ , hence $B=a$ so $f(a)=a^2$ . Hence, for infinitely many positive $a$ , we have $f(a)=a^2$ .

Now, changing $b$ to $f(a)-b$ in the original equation gives $f(b)-bf(2a)+f(a)f(2a)$ is always a perfect square. Now, letting $a$ range over the infinitely many positive integers for which we have shown $f(a)=a^2$ , this becomes $f(b)-4a^2b+4a^4$ is always a perfect square. Now, let $r=f(b)-b^2$ , and observe that $b^2-4a^2b+4a^4=(b-2a^2)$ is always a perfect square, and is always $r$ away from $f(b)-4a^2b+4a^4$ . As $f(b)-4a^2b+4a^4$ can get arbitrarily large (positive), the only square within $r$ away will be $f(b)-4a^2b+4a^4$ itself, so in fact $r=0$ and hence $f(b)=b^2$ for all $b$ . This concludes the proof.

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megarnie

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megarnie

#20 h Mar 2, 2024, 6:13 PM

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The only solutions are $\boxed{f(x) = x^2}$ and $\boxed{ \begin{cases} f(x) = 0 & \text{ if } x \text{ is even } \\ f(x) \text{ is a perfect square } & \text{ if } x \text{ is odd} \\ \end{cases}}$ , where the perfect squares can be arbitrarily chosen. It's easy to check that these work. Now we prove they are the only solutions.

Let $P(a,b)$ denote the given assertion.

$P(a,0): f(f(a))$ is a perfect square.

Claim: $f(0) = 0$ .
Proof: Suppose otherwise.
$P(0,f(0)): f(0) + f(0)^2$ is a perfect square. If $f(0)^2 + f(0) = m^2$ , then $(2f(0) + 1)^2 = (2m)^2 + 1$ , meaning that $(2f(0) + 1 - 2m)(2f(0) + 1 + 2m) = 1$ . Since any two integers multiplying to one are equal we have that $2f(0) + 1 - 2m = 2f(0) + 1 + 2m$ , so $m = 0$ . Since $f(0) \ne 0$ , $f(0) = -1$ must hold.

$P(0,a): f(-a - 1) - a$ is a perfect square, or in other words $f(a) + a + 1$ is a perfect square.
Then $f(f(a)) + f(a) + 1$ and $f(f(a))$ are perfect squares. Since the difference between two squares cannot be $2\pmod 4$ , $f(a)$ cannot be $1\pmod 4$ for any $a$ .

$P(a, f(a)): f(a) f(2a) - 1$ is a perfect square. This implies that if $f(a) \equiv 0\pmod 4$ , then there is a perfect square that is $3\pmod 4$ , absurd. Hence for each $a$ , $f(a) \equiv \{2,3 \}\pmod 4$ .

Consider $f(3)$ . Since $f(3) + 3 + 1$ is a perfect square, $f(3)$ is a quadratic residue mod $4$ , so it must be $0$ or $1$ modulo $4$ , contradiction. Hence $f(0) = -1$ is impossible, so $f(0) = 0$ must hold. $\square$

$P(0,-a)$ implies that $f(a)$ is a perfect square for each $a$ , so also $f(a) \ge 0$ for each $a$ .

$P(a, f(a) - 2a): f(2a)(f(a) - 2a + 1)$ is a perfect square, so either $f(2a) = 0$ or $f(a) - 2a + 1$ is a perfect square.

Case 1: There exists a positive root of $f$ .
Claim: For any positive integer $k$ with $f(k) = 0$ , $f(2k) = 0$ must hold.
Proof: We have $f(2k)(f(k) - 2k + 1)$ is a perfect square, meaning that $f(2k)(1 - 2k)$ is a perfect square. If $f(2k) \ne 0$ , then $f(2k)(1 - 2k) < 0$ (because $f(2k) \ge 0$ ), contradiction. $\square$

$P(a, f(a) - k): (f(a) - k) f(2a)$ is a perfect square for any integer $a$ . If some integer $a$ satisfied $f(2a) \ne 0$ , then $f(a) - k > 0$ must hold. However, by our previous claim, we can replace $k$ with $k \cdot 2^n$ for any positive integer $n$ . Choosing $n$ such that $k \cdot 2^n > f(a)$ gives a contradiction. Thus, $f(2a) = 0$ for all integers $a$ , so $f(x) = 0$ for all even $x$ . Since we had already proven $f(a)$ is a perfect square, this case implies the second solution described in the beginning.

Case 2: There does not exist a positive root of $f$ .
Then for any positive integer $a$ , $f(a) - 2a + 1$ is a perfect square.

Call positive integers $x$ with $f(x) = x^2$ special.

Claim: There are infinitely many special positive integers $x$ .
Proof: Consider any positive integer $x$ where $2x - 1 = p$ is a prime (clearly infinitely many $x$ exist).

Then $f(x) - 2x + 1$ and $f(x)$ are perfect squares with a difference of $2x - 1$ . Notice if $m^2 - n^2 = p$ for some prime $p$ , then $(m-n)(m+n) = p$ hence $(m-n) + (m+n) = (-1) + (-p)$ or $(m-n) + (m+n) = (1 + p)$ , meaning that $m =\pm \frac{p+1}{2}$ .

Since $f(x) - (f(x) - 2x + 1) = 2x -1$ is a prime, we have $f(x) = \left( \frac{2x - 1 + 1}{2} \right) = x^2$ . $\square$

Claim: If $x$ is special, then $2x$ is special.
Proof: If $x$ is special, then $f(x^2 - a) + af(2x)$ is always a perfect square and $x^2 + (f(a) - x) f(2a)$ is also a perfect square.

If $x^2 -a$ is special then $(x^2 - a) ^2 + a f(2x)$ is a square. Let $f(2x) = 4x^2 + c$ . $(2x-1)^2 + c$ is a square

We have $x^4 - 2ax^2 + a^2 + 4x^2 a + a\cdot c$ is a square.

$x^4 + 2ax^2 + a^2 + a\cdot c = (x^2 + a)^2 + a\cdot c$ is a square for infinitely many (negative) $a$ . Hence $P(a) = (a-x^2)^2 - a\cdot c$ is a perfect square for infinitely many positive integers $a> x^2$ .

We have $P(a) = a^2 - (2x^2 + c) a + x^4$ is a perfect square.

If $f(2x)$ was odd, then $f(2x) - 4x + 1\equiv 2\pmod 4$ , so it isn't a perfect square, absurd. Hence $f(2x)$ is even, implying that $c$ is also even. Let $c = 2m$ .

Hence we have $P(a) = a^2 - 2(x^2 + m) a + x^4 = (a - (x^2 + m))^2 - ((x^2 + m)^2 - x^4))$ .

Therefore, $(a - (x^2 + m))^2 - P(a) = (x^2 + m)^2 - x^4$ , so infinitely many distinct squares have a difference of $(x^2 + m)^2 - x^4$ , meaning that $(x^2 + m)^2 = x^4$ , so $m \in \{-2x^2, 0\}$ .

If $m = -2x^2$ , then $f(2x) = 4x^2 - 2m = 0$ , so contradiction since $x$ is a positive integer. Thus, $m = 0$ and $2x$ is also special. $\square$

Now we prove that $f(x) = x^2$ for all integers $x$ . Suppose otherwise. Let $k$ be any integer with $f(k) \ne k^2$ and let $f(k) = k^2 + c$ .

For any special $x$ , $P(x, x^2 - k): k^2 + c + 4x^2 (x^2 - k) = (2x^2 - k)^2 + c$ is a perfect square.

This implies that infinitely many pairs of two squares have a difference of $c$ , so $c=0$ , absurd since $f(k) \ne k^2$ .

Therefore, all integers $x$ satisfy $f(x) = x^2$ and we are done.

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bin_sherlo

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bin_sherlo

#21 h Apr 16, 2025, 9:21 PM • 1 Y

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Answer are $f(a)=a^2$ for all positive integers and $f(2a)=0, \ f(2a+1)\in \mathbb{Z}^2$ . We have $f(b)+(f(a)-b)f(2a)\in \mathbb{Z}^2$ .
Lemma: If $k^2x^2+px+q$ is a perfect square for infinitely many positive integer $x$ , then there exists an integer $r$ such that $2r=p, \ r^2=q$ .

Claim: $f(0)=0$ .
Proof: $P(0,0)$ gives $f(0)^2+f(0)$ is a perfect square hence $f(0)\in \{0,-1\}$ . Suppose that $f(0)=-1$ . Plugging $P(0,b)$ yields $f(b)+b+1\in \mathbb{Z}^2$ . Also $P(a,0)$ implies $f(a)f(2a)-1\in \mathbb{Z}^2$ thus, $4\not | f(a)$ . Hence $f(4a)\equiv -1(mod \ 4)$ . Since $f(2)+3\in \mathbb{Z}^2$ and $f(2)f(4)-1\in \mathbb{Z}^2$ we have $f(2)\equiv 2(mod \ 4)$ . However, $P(2,2)$ gives $f(2)+f(2)f(4)-2f(4)\equiv 2(mod \ 4)$ cannot be a perfect square. Thus, $f(0)=0$ .
Claim: $f(a)$ is perfect square.
Proof: $P(0,b)$ gives the result. Let $f(a)=g(a)^2$ where $g\geq 0$ .
$g(2a)^2g(a)^2-bg(2a)^2+g(b)^2\in \mathbb{Z}^2$ Claim: If $g(m)=0$ for $m\neq 0$ , then $g(2a)=0$ holds.
Proof: Suppose that $g(2l)\neq 0$ . $P(l,m)$ gives $g(2l)^2(g(l)^2-m)\in \mathbb{Z}^2$ or $g(l)^2-m\in \mathbb{Z}^2$ . We observe that $g(m)=0$ implies $m<g(a)^2$ hence $g(x)>0$ for $x>g(a)^2$ . Pick sufficiently large $x$ . Since $g(2x)\neq 0$ , we get $g(x)^2-m\in \mathbb{Z}^2$ and this implies $g(x)<m+10^{15!}$ . By $P(x,2x)$ we also see that $g(x)^2-2x+1\in \mathbb{Z}^2$ however $g(x)-2x+1<0$ .
Since all even integers map to zero, $f(b)\in \mathbb{Z}^2$ sufficies for odds.

Now suppose that $g$ is injective at $0$ .
Claim: $g(\frac{p-1}{2})=\frac{p-1}{2}$ .
Proof: Plugging $P(\frac{p-1}{2},p-1)$ gives $g(\frac{p-1}{2})^2-p\in \mathbb{Z}^2$ and since $x^2-y^2=p$ or $(x-y)(x+y)=p$ implies $x=\frac{p+1}{2}, \ y=\frac{p-1}{2}$ over positive integers, we get $g(\frac{p-1}{2})=\frac{p-1}{2}$ .
Claim: $g(2a)=2g(a)$ .
Proof: $P(a,\frac{p-1}{2})$ implies $(\frac{p-1}{2})^2-g(2a)^2.(\frac{p-1}{2})+g(a)^2g(2a)^2\in \mathbb{Z}^2$ . Hence $x^2-g(2a)^2x+g(a)^2g(2a)^2\in \mathbb{Z}^2$ has infinitely many positive integer solutions. By the lemma, $2|g(a)g(2a)|=g(2a)^2$ or $2g(a)=g(2a)$ .
Claim: $g(a)=|a|\iff f(a)=a^2$ .
Proof: We obtain $4g(a)^4-4g(a)^2b+g(b)^2\in \mathbb{Z}^2$ . Now pick $a=\frac{p-1}{2}$ to see that $4x^2-4bx+g(b)^2\in \mathbb{Z}^2$ has infinitely many solutions over positive integers. Thus, $g(b)=|b|$ or $f(a)=a^2$ as desired. $\blacksquare$

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MathLuis

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MathLuis

#22 h Apr 18, 2025, 3:22 AM

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Oh boi, time to have some real fun and avenge child Luis.
Denote $P(a,b)$ the assertion of the given F.E.
$P(0,f(0))$ gives $f(0)^2+f(0)=k^2$ but then $(2f(0)+1)^2=(2k)^2+1$ which means that since only consecutive squares are $0,1$ we must have that $k=0$ and thus $f(0)=0$ or $f(0)=-1$ .
If $f(0)=-1$ then from $P(a,0)$ we always get (regardless of the case btw) that $f(f(a))$ is a perfect square but also $P(0,-b-1)$ gives that $f(b)+b+1$ is always a perfect square and thus $f(f(a))+f(a)+1$ is always a perfect square so if $f(a)$ was ever $1 \pmod 4$ then we would have $c^2 \equiv f(f(a))+2 \equiv 2,3 \pmod 4$ and of course this is a contradiction, therefore $f(a)$ is not $1 \pmod 4$ at any time which means that $4 \mid f(f(a))$ always holds and thus $4 \mid f(-1)$ and also $f(a)$ will always be $-1,0 \pmod 4$ and thus by $P(a,4k)$ we have that $f(f(a)-4k) \equiv 0 \pmod 4$ and thus $4 \mid f(4\ell)$ for any integer $\ell$ (follows by fixing $a$ ) however now $P(2\ell, b)$ gives $f(f(2\ell)-b)+bf(4\ell)$ is a square congrutn to $f(f(2\ell)-b)) \pmod 4$ and this can be $0,1$ but all $f$ 's are $-1,0$ so since $b$ is any integer we must have $4 \mid f(z)$ for all integers $z$ and of course $4 \not \; \mid f(0)=-1$ thus a contradiction!.
Therefore $f(0)=0$ and from $P(0,a)$ we get that $f(a)$ is always a perfect square.
Now consider $P(a,f(a)-2a)$ which gives that $f(2a)=0$ or $f(a)-2a+1$ is a perfect square, now let $\mathcal D$ the set of numbers with at most two positive divisors then for any $d \in \mathcal D$ that is not $\pm 2$ it happens that $f(d+1)=0$ or $f \left(\frac{d+1}{2} \right)-d$ is a perfect square and in the latter case it means that there exists $k, \ell$ such that $d=k^2-\ell^2$ which leads to $k+\ell=\pm d$ and $k-\ell=\pm 1$ and in either case we would end up having that either $f(d+1)=0$ or $f \left(\frac{d+1}{2} \right)= \left(\frac{d+1}{2} \right)^2$ .
Now the next thing to see is that a difference of squares is always $0 \pmod 4$ or odd this leaves us to consider that if there was a zero $c$ then we would have from $P(a, f(a)-c)$ that $(f(a)-c)f(2a)$ is a perfect square and if $f(a)-c$ was a perfect square then $-c$ is a difference of squares so $c \equiv 2 \pmod 4$ is never going to happen, this also proves that $f(2) \ne 0$ and thus $f(1)=1$ but also for every prime $p \equiv 1 \pmod 4$ we have that $f \left(\frac{p+1}{2} \right)= \left(\frac{p+1}{2} \right)^2$ or else we would have $f(p+1)=0$ in which case it is a contradiction unless we always had $f(2k)=0$ and $f(2k+1)$ a perfect square for all $k \in \mathbb Z$ which indeed works so we will stop considering this solution. And now we also have for all primes $q \equiv 3 \pmod 4$ that $f \left(\frac{1-q}{2} \right)=\left(\frac{1-q}{2} \right)^2$ , and using this and the thing above we will check that if such $c \ne 0$ existed then the sol we said we won't mention anymore will happen again.
Indeed if it weren't the case then we would have that $\left(\frac{p+1}{2} \right)^2-c$ is always a perfect square but setting $p$ large enough we can get its bounded between two squares so we get a contradiction!.
Therefore this means for every $d \in \mathcal D$ we have that $f \left(\frac{d+1}{2} \right)=\left(\frac{d+1}{2} \right)^2$ in this case, the most useful part from this is getting infinitely many $\ell \in \mathbb Z$ such that $f(\ell)=\ell^2$ and plug $P(a,f(a)-\ell)$ to get that $\ell^2-\ell f(2a)+f(a)f(2a)$ is a perfect square and for fixed $a$ this means that by some trivial size argument we have that when $f(a)f(2a)=d^2$ then $f(2a)=2d$ this means $d=2d_1^2$ but also $f(2a)^2=4f(a)f(2a)$ so $f(2a)=4f(a)$ . Thus giving $f(2\ell)=4\ell^2$ .
Also notice from $f(a)-2a+1$ being a perfect square we must have that $0 \le f(a) \le a^2$ for all $a \in \mathbb Z$ by simple size checking and therefore now if we considered $P(\ell, \ell^2-a)$ we get that $f(a)+4(\ell^2-a)\ell^2$ is always a perfect square, sorting this we have $(2\ell^2)^2-2a(2\ell^2)+f(a)$ being a perfect square for infinitely many $\ell$ and thus again from a size argument this implies that $4f(a)=(2a)^2$ and thus $f(a)=a^2$ for all $a \in \mathbb Z$ is the second solution which indeed works, thus we are done :cool:

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