Problem 4

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Problem 4

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cj13609517288

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cj13609517288

#70 h Dec 1, 2022, 4:38 PM

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Setup, using homogeneous

Defining and giving value to g(x) substitution

Casework

Finishing

This post has been edited 8 times. Last edited by cj13609517288, Dec 1, 2022, 4:45 PM

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Taco12

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Taco12

#71 h Dec 21, 2022, 3:58 AM

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writing this up from before because I'm bored

We claim the only solutions are

$f(x)=\begin{cases}0\qquad x \equiv 0 \pmod 2 \\ c\qquad x \equiv 1 \pmod 2\end{cases}$ $f(x)=\begin{cases}0\qquad x\equiv 0\pmod 4 \\ c\qquad x\equiv 1,3\pmod 4 \\ 4c\qquad x\equiv 2\pmod 4\end{cases}$ $f(x)=cx^2.$
Now, we show they are the only ones. Let $P(a,b,c)$ be the assertion. Then,

$P(0,0,0) \rightarrow f(0)=0$ .
$P(a,-a, 0) \rightarrow f(a)=f(-a) \rightarrow f \text{ even }$ .
$P(a,a,-2a) \rightarrow f(2a) = 0 \text{ or } f(2a) = 4f(a)$ .

Claim: If $f(x)=0$ , then $f(cx) = 0$ for all integers $c$ .
Proof. Use induction, with the base case trivial. Assume $f(x)=f(2x)=\dots=f(cx-c)=0$ . Then, $P(x, cx-c, -cx) \rightarrow f(cx)=0. \square$
Now we casework on $f(1), f(2), f(4)$ .

Case 1: $f(2)=0, f(1) \neq 0$ .
By the claim, $f(x)=0$ for even $x$ . Let $n,m$ be odd integers. Then $P(n,m,1-n-m) \rightarrow f(n)=f(m),$ which means $f(x)=c$ for odd $x$ .

Case 2: $f(4)=0, f(1), f(2) \neq 0$ .
We have $f(x)=0$ for $x \equiv 0 \pmod 4$ . Let $n \equiv 1 \pmod 4, m \equiv 3 \pmod 4$ . Then $P(n,m,-n-m) \rightarrow f(n)=f(m),$ which implies $f(x)$ is constant for odd $x$ . Now, let $n\equiv 2 \pmod 4$ . Note that $f(n) \neq 0$ , as otherwise $P(-2, 2-n, n)$ would yield a contradiction. Thus, $f(x)=4x$ for $x \equiv 2 \pmod 4$ .

Case 3: $f(1), f(2), f(4) \neq 0$ .
Note that $f(2) = 4f(1), f(4) = 16f(1)$ . Then, $P(-1, -2, 3) \rightarrow f(3) = f(1) \text{ or } f(3) = 9f(1).$ Similarly, $P(-1,-3,4) \rightarrow f(3) = 9f(1) \text{ or } f(3) = 25f(1),$ and combining gives $f(3)=9f(1)$ . We can use a similar argument inductively to show that $f(x)=cx^2$ .

Case 4: $f(1)=0$ .
This implies $f(x)=0$ which falls under the solution previously found in Case 3.

Since all the solutions work and we have exhausted all cases, we are done. $\blacksquare$

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fclvbfm934

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fclvbfm934

#72 h Oct 21, 2023, 8:13 AM

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Let $P(a, b, c)$ denote the assertion. Quick facts:
$\begin{align*} P(0, 0, 0): f(0) &= 0 \\ P(a, -a, 0): f(a) &= f(-a) \end{align*}$ so we get that $f$ is even. Letting $c = - a - b$ , we have
$f(a)^2 + f(b)^2 + f(a + b)^2 = 2f(a)f(b) + 2f(a+b)[f(a) + f(b)].$ Using the quadratic formula to solve for $f(a + b)$ gives us
$f(a + b) = f(a) + f(b) \pm 2 \sqrt{f(a)f(b)} \qquad{(1)}.$ This tells us a couple of things. The $\sqrt{f(a)f(b)}$ tells us that $f$ must have the same sign everywhere. Furthermore, if we allow $b = 1$ , then this means
$f(a) = \frac{g(a)^2}{f(1)}$ for some $g(a) \ge 0$ , unless $f(1) = 0$ . Note that if $f(1) = 0$ , then by Equation (1), we have $f(a + 1) = f(a)$ , so $f(a) = 0$ everywhere. So assume that $f(1) \neq 0$ .

Replacing all $f$ s with $g$ s in Equation (1) yields
$\begin{align*} g(a + b)^2 &= g(a)^2 + g(b)^2 \pm 2g(a)g(b) \\ \Longrightarrow g(a + b) &= |g(a) \pm g(b)| \qquad{(2)}. \end{align*}$ We know $g(0) = 0$ , and let $g(1) = k$ for some positive integer $k$ . Let $Q(a, b)$ denote plugging in $a$ and $b$ into equation (2).

By $Q(1, 1)$ , we have $g(2) = |k \pm k|$ , so we have two cases.

Case 1: $g(2) = 0$ . Using $Q(x, 2)$ gives us $g(x + 2) = g(x)$ , and so we have $g(x) = 0$ for even $x$ and $g(x) = k$ for odd $x$ .

Case 2: $g(2) = 2k$ . By $Q(2, 1)$ , we have $g(3) = k$ or $g(3) = 3k$ .

Case 2.1: $g(3) = k$ . Then, $Q(3, 1)$ tells us $g(4) = 0$ or $2k$ . But if $g(4) = 2k$ , then $Q(2, 2)$ is a contradiction. So, $g(4) = 0$ , and now by $Q(x, 4)$ , we get that $g$ is periodic with period $4$ , and this determines all of $g$ .

Case 2.2: $g(3) = 3k$ . We shall use induction to show that $g(n) = nk$ for all positive $n$ , with the base case of the first three $n$ handled. Suppose $g(n - 2) = (n-2)k$
and $g(n - 1) = (n-1)k$ . Then we have
$\begin{align*} Q(n - 2, 2): g(n) &= nk \text{ or } (n - 3)k \\ Q(n - 1, 1): g(n) &= nk \text{ or } (n - 2)k \end{align*}$ Notice that there are no absolute value signs here because $n \ge 3$ . The only overlapping value for $g(n)$ is $nk$ , and so we complete our induction.

Solutions: Wrapping up, we have shown that the only admissible solutions are

$g(x) = 0$
$g(x) = k$ for $x \equiv 1 \pmod{2}$ and $0$ otherwise
$g(0) = 0$ , $g(1) = k$ , $g(2) = 2k$ , and $g(3) = k$ , and $g(x) = g(x \pmod{4})$ for all other $x$ .
$g(x) = kx$ for all $x$ .

Then, take $f(x) = c \cdot g(x)$ for any integer $c$ . It is easy to see that all of these are valid solutions.

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BlazingMuddy

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BlazingMuddy

#73 h Dec 5, 2023, 2:25 AM

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There are three classes of solutions to this functional equations; one of them is the nice polynomial function $x \mapsto cx^2$ . The other two classes of solutions can be viewed as a function induced from a solution to the same functional equation but with signatures $\mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}$ , respectively. In general, there is one missing class, induced from a solution with signature $(\mathbb{Z}/2\mathbb{Z})^2 \to \mathbb{Z}$ . Of course, it doesn't appear in this problem since $(\mathbb{Z}/2\mathbb{Z})^2$ is not a group quotient of $\mathbb{Z}$ .

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Let $(G, +)$ be an abelian group and $R$ be an integral domain (with $char(R) \neq 2, 3$ ). Say that $(a, b, c) \in R^3$ is a Heron triple if $a^2 + b^2 + c^2 = 2ab + 2bc + 2ca$ . Find all functions $f : G \to R$ such that $(f(x), f(y), f(z))$ is a Heron triple for any $x, y, z \in G$ with $x + y + z = 0$ .

The case $char(R) = 2$ is not very interesting on itself; the answers are precisely the group homomorphisms.
On the other hand, the case $char(R) = 3$ to be more complicated than the case $char(R) \neq 3$ , since we don't even have $f(0) = 0$ guaranteed anymore!

Answer. A function $f : G \to R$ satisfies the desired properties iff it has one of the following properties:

there exists $c \in R$ non-zero and a group homomorphism $\phi : G \to R$ such that $c f(x) = \phi(x)^2$ for any $x \in G$ ;
there exists $c \in R$ and a surjective group homomorphism $\phi : G \to \mathbb{Z}/2\mathbb{Z}$ such that
$f(x) = \begin{cases} 0, & \phi(x) = 0, \\ c, & \phi(x) = 1; \end{cases}$
there exists $c \in R$ and a surjective group homomorphism $\phi : G \to \mathbb{Z}/4\mathbb{Z}$ such that
$f(x) = \begin{cases} 0, & \phi(x) = 0, \\ c, & \phi(x) = \pm 1, \\ 4c, & \phi(x) = 2; \end{cases}$
there exists a Heron triple and a surjective group homomorphism such that

Since is an integral domain with $char(R) \neq 3$ , for any $r \in R$ , is a Heron triple iff $3r^2 = 0 \iff r = 0$ . In particular, $f(0) = 0$ .
Since is an integral domain, for any $r, s \in R$ , is a Heron triple iff $r = s$ . Together with $f(0) = 0$ , plugging and gives for any $y \in G$ ; i.e., is even. Note that this only requires $f(0) = 0$ , but not $char(R) \neq 3$ . In particular, the original condition is now equivalent to being a Heron triple for any $x, y \in G$ .
Some parts of the solution will assume that $char(R) \neq 3$ , and some parts only assume that $f(0) = 0$ .

First, notice that if $f(x) = 0$ , then is a Heron triple and thus for any $y \in G$ . This means that is a subgroup of $G$ , and that we have a well-defined induced map satisfying for any $x \in G$ , and one can check that is good as well, i.e., is a Heron triple for any $x, y \in G/f^{-1}(0)$ . Thus, we may assume that whenever necessary, as we can just replace with the quotient group $G/f^{-1}(0)$ .

Rearranging the Heron triple condition yields for any $x, y \in G$ . Thus, for any $x, y \in G$ , since and are Heron triples, we get $(f(x + y) - f(x) - f(y))^2 = (f(x - y) - f(x) - f(y))^2 = 4 f(x) f(y)$ , and thus
In particular, this gives for any $x \in G$ . We now divide into three cases:
- Case 1. $f(2x) = 4 f(x)$ for all $x \in G$ .
  
  The corresponding answer is the first category shown above. To motivate the solution, you should think that $f$ is supposed to be similar to the square function.
  
  If $f \equiv 0$ then we are done, so now suppose that $f(g_0) \neq 0$ for some $g_0 \in G$ . If $f$ is the square function, then $f(x + y) - f(x) - f(y)$ would be bilinear in terms of $x$ and $y$ ; it becomes linear if we fix one of $x$ or $y$ . Thus it makes sense to guess that $c = 4 f(g_0)$ and $\phi(x) = f(x + g_0) - f(x) - f(g_0)$ works; it already satisfies $c f(x) = \phi(x)^2$ for all $x \in G$ , and $c \neq 0$ since $f(g_0) \neq 0$ and $4 = 2^2 \neq 0$ in $R$ . It remains to show that $\phi$ is indeed a group homomorphism, which reduces to showing the following equality by substituting $z = g_0$ :
  $f(x + y + z) + f(x) + f(y) + f(z) = f(x + y) + f(y + z) + f(z + x) \quad \forall x, y, z \in G.$
  We first show that $f(x + y) + f(x - y) = 2 f(x) + 2 f(y)$ for any $x, y \in G$ . By (1), either we are done or we can assume $A = f(x + y) = f(x - y)$ . Then $(A, A, f(2x))$ and $(A, A, f(2y))$ are Heron triples, forcing $f(2x) = f(2y) = 4A$ . By the case's hypothesis, this implies $f(x) = f(y) = A$ . But then $(A, A, A) = (f(x), f(y), f(x + y))$ is a Heron triple, which forces $A = 0$ due to $char(R) \neq 3$ . This also yields $f(x + y) + f(x - y) = 2 f(x) + 2 f(y)$ , as desired.
  
  Now the proof of the main equality is as follows: for any $x, y, z \in G$ ,
  $2 f(x + y + z) + 2 f(x) + 2 f(y) + 2 f(z) = f(2x + y + z) + f(y + z) + f(y + z) + f(y - z) = 2 f(x + y) + 2 f(x + z) + 2 f(y + z).$ The second equality holds since $2x + y + z = (x + y) + (x + z)$ and $y - z = (x + y) - (x + z)$ . Now $char(R) \neq 2$ and $R$ is an integral domain, so we can cancel the factor of $2$ out and we are done.
- Case 2. $f(2x) = 0$ for all $x \in G$ .
  
  The corresponding solutions fall into the second and fourth category. Using the quotient $G/f^{-1}(0)$ , we now assume that $f^{-1}(0) = \{0\}$ . In particular, the case's hypothesis implies that $G$ is $2$ -torsion, or equivalently an $\mathbb{F}_2$ -vector space. The claim about what the solutions are is equivalent to saying that $G$ has dimension at most $2$ . It now suffices to show that $\dim_{\mathbb{F}_2}(G) \leq 2$ , i.e., any $3$ elements of $G$ are linearly dependent over $\mathbb{F}_2$ .
  
  Fix some elements $x, y, z \in G$ ; the goal is to show that one of $x, y, z$ , $x + y, x + z, y + z$ , or $x + y + z$ equals zero. Recall that $a^2 + b^2 + c^2 = 2ab + 2bc + 2ca$ can be rearranged to $(c - a - b)^2 = 4ab$ . Then $4 f(x) f(w)$ is a square for each $w \in \{x, y, z, x + y, x + z, y + z, x + y + z\}$ . For each $\alpha, \beta, \gamma \in \{0, 1\}$ (i.e. $\mathbb{F}_2$ ), pick some $r_{\alpha \beta \gamma} \in R$ such that $4 f(x) f(\alpha x + \beta y + \gamma z) = r_{\alpha \beta \gamma}^2$ . We can change $r_{\alpha \beta \gamma}$ to $-r_{\alpha \beta \gamma}$ if desired.
  
  Since $(f(x), f(y), f(x + y))$ is Heron, after multiplying by $(4 f(x))^2$ , we get that $(r_{100}^2, r_{010}^2, r_{110}^2)$ is also Heron, which forces $r_{110} = r_{100} \pm r_{010}$ . The same method yields $r_{\mathbf{v}_1 + \mathbf{v}_2} = r_{\mathbf{v}_1} \pm r_{\mathbf{v}_2}$ for each $\mathbf{v}_1, \mathbf{v}_2 \in \mathbb{F}_2^3$ . Fix a sign of $r_{100}$ , then choose a sign for $r_{010}$ such that $r_{110} = r_{100} + r_{010}$ , and then choose a sign for $r_{001}$ such that $r_{111} = r_{110} + r_{001} = r_{100} + r_{010} + r_{001}$ . Now we list some values of $f$ :
  $\begin{align*} f(x) &= r_{100}^2 \\ f(y) &= r_{010}^2 \\ f(z) &= r_{001}^2 \\ f(x + y) &= (r_{100} + r_{010})^2 \\ f(x + z) &= (r_{100} \pm r_{001})^2 \\ f(y + z) &= (r_{010} \pm r_{001})^2 \\ f(x + y + z) &= (r_{100} + r_{010} + r_{001})^2 \end{align*}$ Recall that the goal is to show that at least one of the entries above is equal to $0$ . Suppose for the sake of contradiction that none of them are zero. We just case-bash based on the choices of sign in the two $\pm$ s.
  
  First, since $(f(x), f(y + z), f(x + y + z))$ is Heron, we get that $f(y + z)$ is equal to either $(r_{010} + r_{001})^2$ or $(2 r_{100} + r_{010} + r_{001})^2$ . In the latter case, some choices of sign of $\pm$ yields $2 r_{100} + r_{010} + r_{001} = \pm r_{010} \pm r_{001}$ . If both signs are $-$ , then $r_{100} + r_{010} + r_{001} = 0$ and thus $f(x + y + z) = 0$ . If both signs are $+$ , then $r_{100} = 0$ . If one of the sign is $-$ , say $2 r_{100} + r_{010} + r_{001} = -r_{010} + r_{001}$ , then $r_{100} + r_{010} = 0$ and thus $f(x + y) = 0$ ; so $f(y + z) = (2 r_{100} + r_{010} + r_{001})^2$ yields a contradiction. The only case remaining is that $f(y + z) = (r_{010} + r_{001})^2$ . By symmetry, avoiding contradiction also yields $f(x + z) = (r_{100} + r_{001})^2$ .
  
  Now consider that $(f(x + y), f(x + z), f(y + z))$ is Heron. This implies that, for some choice of the signs $\pm$ ,
  $\pm (r_{100} + r_{010}) \pm (r_{100} + r_{001}) \pm (r_{010} + r_{001}) = 0.$ WLOG suppose that at least two of the signs are $+$ . If all three are equal to $+$ , then we get $r_{100} + r_{010} + r_{001} = 0$ and thus $f(x + y + z) = 0$ . Otherwise, WLOG the signs for $r_{100} + r_{010}$ and $r_{100} + r_{001}$ are $+$ . Then the previous equality becomes $2 r_{100} = 0$ , which yields that $r_{100} = 0$ and thus $f(x) = 0$ ; a contradiction as well. All cases yield a contradiction, so one of the seven values shown above must be equal to $0$ . We are done.
  
  (There should be an alternative way, free of the above kind of bash, but either I didn't actually do it or that I forgot how to do it.)
- Case 3. There exists $g \in G$ such that $f(2g) = 4 f(g) \neq 0$ and $g' \in G$ such that $f(g') \neq 0$ and $f(2g') = 0$ .
  
  Again, pass on to the quotient and WLOG assume that $f^{-1}(0) = \{0\}$ . The corresponding solutions fall into the third category; we claim that $g$ is a generator of $G$ of order $4$ , which also implies that $g' = 2g$ .
  
  For convenience, let $f(g) = A \neq 0$ . Since $2g' = 0$ , we have $f(g + g') = f(g - g') = f(g' - g)$ . On the other hand, $(f(g + g'), f(g' - g), f(2g) = 4A)$ is a Heron triple, so we necessarily have $f(g + g') = f(g' - g) = A$ . Since $f(g') \neq 0$ and $(A, A, f(g')) = (f(g), f(g' - g), f(g'))$ is Heron, we get $f(g') = 4A$ . Now both $(f(g' - 2g), f(g' - g), f(g)) = (f(g' - 2g), A, A)$ and $(f(g' - 2g), f(g'), f(2g)) = (f(g' - 2g), 4A, 4A)$ are Heron, so either $f(g' - 2g) = 0$ or $f(g' - 2g) = 4A = 16A$ . Since $char(R) \neq 2, 3$ , the latter yields $A = 0$ ; a contradiction. Thus $f(g' - 2g) = 0 \implies g' = 2g$ , and so $4g = 2g' = 0$ .
  
  Fix any $x \in G$ ; the goal is to show that $x \in \{0, g, 2g, -g\}$ . By (1), we get either $f(x + 2g) = f(x)$ or $f(x + 2g) + f(x) = 2 f(x + g) + 2 f(g)$ . Similarly, either $f(x + 2g) = f(x)$ or $f(x + 2g) + f(x) = 2 f(x - g) + 2 f(g)$ holds, so we have either $f(x + 2g) = f(x)$ or $2 f(x + g) + 2 f(g) = 2 f(x - g) + 2 f(g) \implies f(x + g) = f(x - g)$ . It now suffices to show that for any $x \in G$ , $f(x + g) = f(x - g)$ implies $x \in \{0, 2g\}$ ; then we can apply the result to either $x$ or $x + g$ .
  
  Fix any $x \in G$ such that $f(x + g) = f(x - g)$ . Since $(f(x + g), f(x - g), f(2g)) = (f(x + g), f(x + g), 4A)$ is Heron and $A \neq 0$ , we get $f(x + g) = A = f(g)$ . Then we get $f(x + 2g), f(x) \in \{0, 4A\}$ . If $f(x + 2g) = f(x) = 4A = f(2g)$ , then $(4A, 4A, 4A)$ is a Heron triple; a contradiction, since $4A \neq 0$ . So either $f(x + 2g) = 0 \implies x + 2g = 0 \implies x = 2g$ , or $f(x) = 0 \implies x = 0$ .

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megarnie

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megarnie

#74 h Jan 28, 2024, 3:33 PM

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Writing this up again I forgot my solution anyway:

The only solutions are $f\equiv 0$ , $f(x) = cx^2$ , $f(x) = \begin{cases} 0 & x\equiv 0\pmod 2 \\ c & x\equiv 1\pmod 2 \\ \end{cases}$ , and $f(x) = \frac{n + |n|}{2} \cdot c$ , for any integer constant $c\ne 0$ , where $x_4$ is the remainder when $x$ is divided by $4$ and $n = \nu_2( 2^{x_4} + x_4^3 + x_4^2 + 8 x_4 ) - 1$ . We can check that these work, now we prove they are the only solutions.

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

$P(0,0,0): f(0) = 0$ .

Claim: If $a + b + c = 0$ and $f(c) = 0$ , then $f(a) = f(b)$ .
Proof: We have $f(a)^2 + f(b)^2 = 2f(a)f(b)$ , so $(f(a) - f(b))^2 = 0\implies f(a) = f(b)$ . $\square$

This implies that $f$ is even.

$P(a, a, -2a): 2f(a)^2 + f(2a)^2 = 2f(a)^2 + 4f(a)f(2a)$ , so either $f(2a) = 0$ or $f(2a) = 4f(a)$ for each $a$ .

The equation implies that $f(a) ^2 + f(b)^2 + f(a+b)^2 = 2f(a)f(b) + 2f(b) f(a+b) + 2f(a+b)f(a) ,$ so $f(a+b)^2 - (2(f(a) + f(b))) f(a+b) + (f(a) - f(b))^2 = 0,$ so using the quadratic formula, $f(a+b) = \frac{ 2f(a) + 2f(b) \pm \sqrt{4 ( (f(a) + f(b))^2 - (f(a) - f(b))^2) }}{2} = \frac{2f(a) + 2f(b) \pm 4\sqrt{f(a)f(b)}}{2}$
This can be simplified to $f(a) + f(b) \pm 2\sqrt{f(a)f(b)} = \left(\sqrt{f(a)} \pm \sqrt{f(b)}\right)^2$ . Let $Q(a,b)$ denote the assertion that $f(a+b) = \left(\sqrt{f(a)} \pm \sqrt{f(b)}\right)^2$
Note that if $f(x) = 0$ for some $x$ , then $f(a + x) = f(a)$ by $P(a,x)$ , so $f(a) = f( a\pmod x )$ (in fact implying that $f(kx) = 0$ for any integer $k$ ).

Thus if $f(1) =0$ , then $f\equiv 0$ . So assume $f(1) = c \ne 0$ .

Now if $f(2) = 0$ , we see that $f(2x) = 0$ for any integer $x$ , and $f(2x+ 1) = c$ due to $f$ being periodic with period $2$ (from $P(a,2)$ ). This corresponds to one of the solutions described in the beginning. Now we may assume $f(2) \ne 0$ , so $f(2) = 4c$ .

If $f(4) = 0$ , then $f(4x) = 0$ for any integer $x$ , so $f$ is periodic with period $4$ . Since $-1 \equiv 3 \pmod 4$ and $f$ is even, we have $f(1\pmod 4) = f(3\pmod 4 )= c$ and $f(2\pmod 4) = 4c$ , corresponding to one of the solutions in the beginning.

Now assume $f(1), f(2), f(4)$ are nonzero. We have $f(1) = c, f(2) = 4c, f(4) = 16c$ . From $P(2, 1)$ , we see that $f(3)$ is either $\left(\sqrt{c} + 2\sqrt{c}\right)^2 = 9c$ or $(\sqrt{c} - 2\sqrt{c})^2 = c$ . If $f(3) = c$ , from $P(3,1)$ , we see that $16c = \left(\sqrt{c} \pm \sqrt{c}\right)^2$ , but this means $c = 0$ , absurd. Thus, $f(3) = 9c$ . Now we induct to show $f(n) = n^2 c$ for any positive integer $n$ , with base cases $1,2,3,4$ .

Suppose it's true for $1,2,\ldots, n$ and $n\ge 4$ . We have from $P(n,1)$ that $f(n+1) = \left( (n \pm 1) \sqrt{c} \right)^2 = (n \pm 1)^2 c$ . If $f(n+1) = (n-1)^2 c$ , then $P(n-1,2)$ gives that $f(n +1) = \left( (n-1) \sqrt{c} \pm 2 \sqrt{c} \right)^2$ , which is either $(n-3)^2 c$ or $(n+1)^2 c$ , but this means that $(n-1)^2 = (n-3)^2$ or $(n+1)^2 = (n+3)^2$ , both are false for $n > 2$ . Thus, we must have $f(n+1) = (n+1)^2 c$ , and the induction is complete.

Then of course $f(n) = n^2 c$ for any integer $n$ because $f$ is even.

This post has been edited 3 times. Last edited by megarnie, Jan 28, 2024, 3:40 PM

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naonaoaz

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naonaoaz

#75 h Feb 12, 2024, 3:00 PM

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Let $P(x,y,z)$ be the assertion. $P(0,0,0) \implies f(0) = 0$ . Let $b = 0$ to get
$f(a)^2+f(-a)^2 = 2f(a)f(-a) \implies (f(a)-f(-a))^2 = 0 \implies f(a) = f(-a)$ for all integers $a$ . Next letting $c = -(a+b)$ gives
$f(a+b)^2 - 2(f(a)+f(b))f(a+b) + (f(a)-f(b))^2 = 0$ $\implies f(a+b) = f(a)+f(b) \pm 2\sqrt{f(a)f(b)}$ Thus either all $f(x) \ge 0$ or all $\le 0$ or else there will be an complex output of $f$ . Since there is a bijection between all positive $f(x)$ and negative, WLOG all $f(x) \ge 0$ . Thus
$\sqrt{f(a+b)} = \left(\sqrt{f(a)}\pm \sqrt{f(b)}\right)^2 \implies g(a+b) = g(a) \pm g(b)$ where $g(x) = \sqrt{f(x)}$ . Note that $g(2) = g(1) \pm g(1)$ .

If $g(2) = 0$ , then we can inductively show $g(2k) = 0$ for all integer $k$ . Next notice $g(2k+1) = g(2k) \pm g(1) = g(1)$ where we took $+$ sign to avoid $g(2k)$ being negative.
If $g(2) = 2g(1)$ . Next if $g(4) = 0$ , then $g(4k) = 0$ for all integer $k$ . Next $g(4k+1) = g(4k) \pm g(1) = g(1)$ . Similarly, $g(4k+2) = 2g(1)$ . Lastly since $0 = g(1) \pm g(4k+3)$ , so $g(4k+3) = g(1)$ .

If $g(4) = 4g(1)$ , then $g(3) = 3g(1)$ . Next since $g(5) = 4g(1) \pm g(1) = 3g(1) \pm 2g(1)$ , we must have $g(5) = 5g(1)$ . Induction gives $g(n) = n g(1)$ .

Concluding, we see

$f(x) = 0$ for even $x$ , and $f(x) = c$ for odd $x$ .
$f(x) = 0$ for $x \equiv 0 \pmod 4$ , $f(x) = c$ for $x \equiv 1,3 \pmod 4$ , and $f(x) = 4c$ if $x \equiv 2 \pmod 4$ .
$f(x) = cx^2$ for all $x$ .

for any $c \in \mathbb{Z}$ .

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AngeloChu

471 posts

AngeloChu

#76 h Mar 4, 2024, 2:34 AM

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only a simple sketch of my full sol since i got lazy

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KevinYang2.71

428 posts

KevinYang2.71

#77 h May 7, 2024, 7:45 PM

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We claim the only functions are $\boxed{f(x)\equiv cx^2}$ ,
$\boxed{f(x)\equiv \begin{cases} 0 & \text{if $x\equiv 0\pmod 2$}\\ c & \text{if $x\equiv 1\pmod 2$} \end{cases} },$ and
$\boxed{f(x)\equiv \begin{cases} 0 & \text{if $x\equiv 0\pmod 4$}\\ c & \text{if $x\equiv 1\pmod 2$}\\ 4c & \text{if $x\equiv 2\pmod 4$} \end{cases} }$ for a constant $c\in\mathbb{Z}$ . It is not hard to check that these work.

Let $P(a,b,c)$ denote the given assertion. From $P(0,0,0)$ we get $f(0)=0$ and $P(a,-a,0)$ gives us $f(a)=f(-a)$ . Then $P(a,b,-a-b)$ yields
$f(a)^2+f(b)^2+f(a+b)^2=2f(a)f(b)+2f(a+b)(f(a)+f(b))$ so
$f(a+b)=f(a)+f(b)\pm\sqrt{2f(a)f(b)}\tag{*}$ for all $a,b\in\mathbb{Z}$ . Since $f(a+1)$ is real, it follows that $f(a)$ and $f(1)$ have the same sign. If $f(1)$ is negative we can multiply $(*)$ by $-1$ so $-f$ is positive and we can simply let $c$ be negative in our boxed functions. Thus we may WLOG assume $f(1)\geq 0$ . Then $\sqrt{f(a+b)}=\left|\sqrt{f(a)}\pm\sqrt{f(b)}\right|$ . Let $g(x):=\sqrt{f(x)}$ so $g(a+b)=|g(a)\pm g(b)|$ for all $a,b\in\mathbb{Z}$ . Note that $g(n)=g(-n)\geq 0$ for all $n\in\mathbb{Z}$ . Let $r:=g(1)$ . Then $g(2)\in\{0,2r\}$ .

Case 1: $g(2)=0$ . We prove by induction on $|x|$ that
$g(x)\equiv \begin{cases} 0 & \text{if $x\equiv 0\pmod 2$}\\ r & \text{if $x\equiv 1\pmod 2$} \end{cases}$ with the base cases trivial. Assume the statement for $|x|\leq k$ (with $k\geq 2$ ). Then $g(k+1)=|g(k-1)\pm g(2)|=g(k-1)$ so the induction step is complete. This yields the second solution set with $c:=r^2$ .

Case 2: $g(2)=2r$ . Then $g(3)\in\{r,3r\}$ .

Case 2.1: $g(3)=r$ . We prove by induction on $|x|$ that
$g(x)\equiv \begin{cases} 0 & \text{if $x\equiv 0\pmod 4$}\\ r & \text{if $x\equiv 1\pmod 2$}\\ 2r & \text{if $x\equiv 2\pmod 4$} \end{cases}$ with the base cases trivial. Note that $g(4)=|g(3)\pm g(1)|\in\{0,2r\}$ and $g(4)=|g(2)\pm g(2)|\in\{0,4r\}$ so $g(4)=0$ . Assume the statement for $|x|\leq k$ (with $k\geq 4$ ). Then $g(k+1)=|g(k-3)\pm g(4)|=g(k-3)$ so the induction step is complete. This yields the third solution set with $c:=r^2$ .

Case 2.2: $g(3)=3r$ . We prove by induction on $|x|$ that $g(x)\equiv rx$ with the base cases trivial. Assume the statement for $|x|\leq k$ (with $k\geq 3$ ). Then $g(k+1)=|g(k)\pm g(1)|=\{c(k-1),c(k+1)\}$ and $g(k+1)=|g(k-1)\pm g(2)|=\{r(k-3),r(k+1)\}$ so $g(k+1)=r(k+1)$ . This yields the first solution set with $c:=r^2$ .

We are done. $\square$

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ezpotd

1286 posts

ezpotd

#78 h Sep 7, 2024, 5:59 AM

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I claim the solution set is given by $f = kx^2$ , $f = 0$ , $f$ giving distinct values mod $2$ , and lastly $f$ giving values mod $4$ , where it gives $0$ for $0$ , $k$ for $1$ , $4k$ for $2$ , $k$ for $3$ . We check each of these work.

For the first function, we desire to prove $a^4 + b^4 + c^4 = 2a^2b^2 + 2b^2c^2 + 2a^2c^2$ , setting $c= -(a + b)$ and raw expansion finishes.
The second function is obvious.
The third function just requires us to analyze the summation cases mod $2$ , of which there are only two, $0 + 0 + 0 =0$ and $0 + 1 + 1 = 0$ , both of which obviously work.
The last function requires us to analyze the summation cases mod $4$ , of which there are just three, $0 + 0 + 0 = 0, 0 + 1 + 3 = 0, 0 + 2 + 2 = 0$ , all of which obviously work.

Set $a,b,c = 0$ to get $f(0) = 0$ . Then set $a = -b, c = 0$ to get $f(a) = f(-a)$ . We use this without mention, effectively restraining the problem over positive values. Now take $a = b, c = -2a$ , and we get $2f(a)^2 + f(-2a)^2 = 2f(a)^2 + 4f(a)f(-2a)$ , giving $f(-2a)^2 = 4f(a)f(-2a)$ , or $f(2a)(f(2a) - 4f(a)) = 0$ . Thus for all $a$ , we have $f(2a) = 0$ OR $f(2a) = 4f(a)$ . We now divide into two cases.

Case 1: $f(n) = 0$ for some positive $n$ . We first inductively prove $f(kn )= 0$ for all $k$ . The case of $k = 2$ is obvious. Then take $a = n, b = kn, c = -kn - n$ , we get $f(kn + n)^2 = 0$ as desired. Now we take $n | x + y$ , setting $a = x ,b = y, c = -n$ gives $f(x)^2 + f(y)^2 = 2f(x)f(y)$ , giving $f(x) = f(y)$ . This means that if $x \equiv -y \mod n, f(x) = f(y)$ (this also implies that $f$ operates modulo $n$ ). Now we take the smallest positive root of $f$ , $n$ , and we claim that all other roots are multiples of $n$ . Assume this is not the case, then there must be another root $m$ not divisible by $n$ , and $f(n - (m \mod n))$ must be equal to $f(m)$ and thus equal to $0$ , forcing a smaller multiple, contradiction. It suffices now to do casework on $n$ . If $n = 1$ , we get $f$ is just $0$ over its entire domain. If $n$ has an odd prime divisor, we can find some solution with $k > 0, n \nmid x$ to $2^k x \equiv x \mod n$ , which gives $f(2^kx) = 4f(2^{k - 1}x) = \cdots = 2^{2k}f(x)$ , but since $f$ operates modulo $n$ , we must have $f(x) = 2^{2k}f(x)$ , which has no solutions since $f(x) \neq 0$ . If $n = 2$ , the only possible solution is $f$ has some value over odds and another over evens, which always works. If $n = 4$ , we must have $f(1) = f(3), f(2) \neq 0$ , the latter of which forces $f(2) = 4f(1)$ . This constrains the solution to a mod $4$ repetition of $f(0) = 0, f(1) = k, f(2) = 4k, f(3) = k$ , which always works. If $n \ge 8$ , we have $f(\frac n2) > 0$ , so $\frac 14(n^2)f(3) = \frac{1}{16}n^2f(6) = \cdots f(\frac n2) = \cdots = \frac{1}{16}n^2 f(2) = \frac 14 n^2 f(1)$ , so $f(1) = f(3)$ , then taking $a = 1, b = 3, c= -4$ , gives $f(4) = 4f(1)$ , contradiction.

Case 2: $f(n) \neq 0$ for all positive $n$ . In this case take $f(1) = k$ , now we inductively prove $f(n) = cn^2$ for positive integers, then using $f$ even finishes. The base case of $n = 1,2$ are trivial. After this, take $a = 1, b = n, c= -n - 1$ , this gives $c^2 + c^2n^4 + f(n + 1)^2 = 2c^2n^2 + (2cn^2 + 2c)f(n + 1)$ , which gives $f(n + 1) = c(n + 1)^2, c(n - 1)^2$ . If the latter is true, substitute $a = n , b =n , c = -2n$ , we then get $f(2n) = 4f(n) = 4cn^2$ , but substituting $a = n - 1, b = n+ 1 , c = -2n$ gives $f(2n) = 4f(n - 1) = 4c(n - 1)^2$ , contradiction. This forces $f(n + 1) = c(n + 1)^2$ .

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pie854

243 posts

pie854

#79 h Sep 12, 2024, 12:17 PM

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Let $P(a,b,c)$ denote the given assertion. Then, $\begin{align*} P(0,0,0) & \ \Rightarrow 3f(0)^2=6f(0)^2 \Rightarrow f(0)=0 \\ P(a,-a,0) & \ \Rightarrow (f(a)-f(-a))^2=0 \Rightarrow f(a)=f(-a). \end{align*}$ So $f$ is even, we'll use this fact a lot now. If $f(1)=0$ then $P(n+1,n,-1)$ gives us $f(n+1)=f(n)$ for all $n$ so by induction $f(n)=0$ . This clearly works.

Let us now assume $f(1)$ is not $0$ , then by scaling lets assume $f(1)=1$ . $P(1,1,-2)$ implies that $f(2)\in \{0,4\}$ . Let us first consider $f(2)=0$ then it's easy to find $f(n+2)=f(n)$ for all $n$ . So $f (n)=\begin{cases} 0 & n\equiv 0 \pmod 2 \\ 1 & n\equiv 1 \pmod 2\end{cases}.$ It's not too hard to verify by casework that this is indeed a solution.

Let's consider $f(2)=4$ . Suppose $f(n)=n^2$ for all $0 \leq n\leq k$ for $k\geq 2$ . Then $P(k,1,-k-1) \Rightarrow f(k+1)^2-2f(k+1)(k^2+1)+(k^2-1)^2=0 \Rightarrow f(k+1)=(k\pm 1)^2.$ Suppose it were that $f(k+1)=(k- 1)^2$ . Then $f(k-1)=f(k+1)$ and $\begin{align*} P(k+1,k-1,-2k) & \ \Rightarrow f(2k)^2=4f(2k)f(k+1) \\ P(k,k,-2k) & \ \Rightarrow f(2k)^2=4f(2k)f(k) \end{align*}$ This implies $f(2k)=0$ . So, $P(n,2k,-n-2k)$ gives us $f(n+2k)=f(n)$ . We have $P(k-1,k-1,-2k+2)\Rightarrow 2(k-1)^4+2^4=4(k-1)^2\cdot 2^2+2\cdot (k-1)^2\cdot (k-1)^2 \Rightarrow k=0,2 \qquad (\star)$ so $k=2$ . So, $f(n+4)=f(n)$ and by induction we get $f (n)=\begin{cases} 0 & n\equiv 0 \pmod 4 \\ 1 & n\equiv 1,3 \pmod 4 \\ 4 & n\equiv 2 \pmod 4\end{cases}.$ Again, it's not too hard to verify by casework that this is indeed a solution. Now if $f(k+1)=(k+1)^2$ then by induction $f(n)=n^2$ for all $n$ , which clearly works. Note that the induction works because $k+1>2$ - look at $(\star)$ .

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cursed_tangent1434

639 posts

cursed_tangent1434

#80 h Dec 21, 2024, 3:03 PM

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Pretty nasty casework. Note that,
$P(0,0,0)\implies 3f(0)^2=6f(0)^2$ which gives us that $f(0)=0$ . Now, $P(a,-a,-0)$ gives us that
$\begin{align*} f(a)^2+f(-a)^2 &= 2f(a)f(-a)\\ f(a)^2-2f(a)f(-a)+f(-a)^2 &= 0\\ (f(a)-f(-a))^2 &= 0 \end{align*}$ Thus, $f(-a)=f(a)$ for all $a \in \mathbb{Z}^+$ . Now, look at $P(a,b,-a-b)$ to obtain that,
$f(a)^2+f(b)^2+f(-a-b)^2 =2f(a)f(b)+2f(b)f(-a-b)+2f(-a-b)f(a)$ So,
$\begin{align*} f(a+b)^2-2f(a+b)(f(a)+f(b))+f(a)^2-2f(a)f(b)+f(b)^2&=0 \\ (f(a+b)^2-f(a)-f(b))^2 &= 4f(a)f(b) \end{align*}$
Now, note that, this means,
$(f(2)-2f(1))^2=4f(1)^2$ and thus, $f(2)=4f(1)$ or $f(2)=0$ . Based on these two value, we have several cases.

Case 1 : $f(1)=c$ and $f(2)=4c$ . In this case note that we have,
$(f(3)-f(2)-f(1))^2=4f(2)f(1)=16f(1)^2$ so, $f(3)=9c$ or $f(3)=c$ . So, we have two sub cases.

\textbf{Subcase 1 : } $f(3)=9c$ . Here, simply note that if we have $f(1)=c,\dots, f(n)=n^2c$ for some $n\geq 3$ then,
$\begin{align*} (f(n+1)-f(n)-f(1))^2 &= 4f(n)f(1)\\ (f(n+1)-n^2c-c)^2 &= 4n^2c^2 \end{align*}$ So, $f(n+1)=(n-1)^2c$ or $f(n+1)=(n+2)^2c$ . But if $f(n+1)=(n-1)^2c$ ,
$\begin{align*} (f(n+1)-f(n-1)+f(2))^2 &= 4f(n-1)f(2)\\ (f(n+1)-(n-1)^2c+4c)^2 &= 16(n-1)^2c^2\\ 16c^2 &= 16(n-1)^2c^2 \end{align*}$ which is clearly impossible for all $n\geq 3$ . Thus, we must have $f(n+1)=(n+1)^2c$ . Thus, by the PMI, we have that for all integers $k \geq 1$ , $f(k)=k^2f(1)$ . This gives us the function $f(n)=n^2c$ for some constant $c$ for all positive integers $n$ .

Subcase 2 : $f(3)=c$ . In this case note that we have,
$(f(4)-8c)^2=(f(4)-2f(2))^2=4f(2)^2=64c^2$ So, $f(4)=16c$ or $f(4)=0$ . But note that if $f(4)=16c$ we have
$4c^2=4f(3)f(1)=(f(4)-f(3)-f(1))^2=(16c-c-c)^2$ so, $4c^2=196c^2$ which requires, $c=0$ , which implies that $f(4)=0$ anyways. Thus, in all cases we have that $f(4)=0$ .

Note that then, we can obtain
$\begin{align*} (f(n+4)-f(n)-f(4))^2 &= 4f(n)f(4)\\ f(n+4)-f(n)-f(4) &= 0 \\ f(n+4) &= f(n) \end{align*}$ for all $n \geq 1$ . Thus, the function is periodic with a period of $4$ giving us,
$f(n)= \begin{cases} c & \text{if } n \equiv 1,3 \pmod{4}\\ 4c & \text{if } n \equiv 2 \pmod{4}\\ 0 & \text{if } n \equiv 0 \pmod{4} \end{cases}$ for all $n \in \mathbb{N}$ for some constant $c \in \mathbb{N}$ .

Case 2 : $f(2)=0$ . Here, note that
$\begin{align*} (f(n+2)-f(n)-f(2))^2 &= 4f(n)f(2)\\ f(n+2)-f(n)-f(2) &= 0\\ f(n+2) &= f(n) \end{align*}$ for all $n\geq 1$ . Thus, the function is periodic with a period of $2$ giving us the function,
$f(n)= \begin{cases} c & \text{if } n \equiv 1 \pmod{2}\\ 0 & \text{if } n \equiv 0 \pmod{2} \end{cases}$ Thus, we have exhausted all cases and are done.

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AshAuktober

1009 posts

AshAuktober

#82 h Jan 23, 2025, 7:45 AM

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Nice problem!

Show that $f(0) = 0, f(-x) = f(x)$ .
Then from here, assuming $f(1) \ne 0$ , show that if $f(n) = n^2f(1)$ , then $f(n+1) \in \{(n-1)^2f(1), (n+1)^2f(1)\}$ .
From here, if $f(n+1) = (n+1)^2f(1) \forall n,$ then $f(x) = f(1)x^2$ . Further note that if $f(2) = (1-1)f(1)^2 = 0$ then we get the solution $f(x) = \begin{cases} f(1), \quad \text{ if } n \equiv 1 \pmod{2}\\ 0, \quad \text{ if } 2 \mid n. \end{cases}$ so assume otherwise.
If $f(n+1) = (n-1)^2f(1)$ for some $n \in \mathbb{N}_{\ge 2}$ , with $f(x) = x^2f(1) \forall x \le n$ , comparing the equations of $(2, n-1, a)$ and $(0, n+1,a)$ gives us $f(2)^2 = 4f(2)(n-1)^2f(1)$ , or $f(2) \in \{0, 4(n-1)^2f(1)\}$ . $f(2) \ne 0$ , so $f(2) = 4(n-1)^2f(1)$ .
But $4f(1) = f(2) = 4(n-1)^2f(1) \implies n \in \{0, 2\}$ . $n = 0$ is false as $n \in \mathbb{N}_{\ge 2}$ . So consider $n = 2$ . This is basically the same as $f(3) = f(1)$ . Further we may assume $f(2) = 4f(1)$ as previous.
Then $f(4) = f(2\times 2) \in \{0, 16f(1)\},$ and $f(4) = f(3+1) \in \{0, 4f(1)\}$ . Since $f(1) \ne 0$ , this implies $f(4) = 0$ . In particular, $f$ is periodic with period 4, so we get
$f(x) = \begin{cases} 0, \quad \text{ if } 4 \mid n.\\ f(1) \quad \text{ if } n \equiv 1 \pmod{2}.\\ 4f(1) \quad \text{ if } n \equiv 2 \pmod{4}. \end{cases}$ All of the functions extracted above work, so we're done. $\square$

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Ilikeminecraft

658 posts

Ilikeminecraft

#83 h Mar 6, 2025, 2:35 AM

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what the impossible p4

The answer is:
$f\equiv\begin{cases}0&x\equiv0\bmod4\\k&x\equiv1,3\bmod4\\4k&x\equiv2\bmod4\end{cases}$ $f\equiv\begin{cases}0&x\equiv0\bmod2\\k&x\equiv1\bmod2\end{cases}$ $f\equiv 0$
Begin by taking $(0, 0, 0) \implies f(0) = 0, (0, a, -a) \implies f(a)$ is even.

Observe that $(f(a) + f(b) - f(c))^2 = f(a)^2 + f(b)^2 + f(c)^2 - 2f(c)f(a) - 2f(c)f(b) + 2f(a)f(b) = 4f(a)f(b).$ Hence, $f(a)f(b)$ is a perfect square. Let $f \equiv k g^2,$ where $k\in\mathbb Z, g\colon\mathbb Z\to\mathbb N_0$ and $g$ is even. Hence, $g(a)^2 + g(b)^2 - g(a + b)^2 = \pm 2g(a)g(b).$ Hence, $g(a + b) = g(a) + g(b)$ or $g(a + b) = |g(a) - g(b)|.$ Hence, we can assume that $g(1) = 1$ .

Now, assume there exists $k$ such that $g(k) = 0.$ Then $g(a) = g(k - a) = 0$ or $g(a) = g(k - a),$ and thus, if $g(k) = 0,$ then $g(a) = g(k - a) = g(a - k).$ Hence, $g$ has period $k.$ However, we also know $g$ is even, and thus, if $k$ is odd, then $g\equiv 0.$ Otherwise, the period is even. If $k= 2, 4,$ then $g\equiv x\pmod2$ or
$g\equiv\begin{cases}0&x\equiv0\bmod4\\1&x\equiv1,3\bmod4\\2&x\equiv2\bmod4\end{cases}$ Now, assume $k> 4.$ In the original equation, take $(a, a, 2a)$ to get $f(2a) = 4f(a)$ or $f(2a) = 0.$ However, this implies $k \leq 4.$

Now, assume that $g$ has no other roots. Now, we induct. Assume there was a pair $(a, b)$ such that $g(a + b) = |g(a) - g(b)| < g(a) + g(b) = g(a) + g(b - 1) + 1 = g(a + b - 1) + 1,$ which finishes. Thus, $g\equiv x.$

This post has been edited 2 times. Last edited by Ilikeminecraft, Mar 29, 2025, 11:29 PM

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Nari_Tom

117 posts

Nari_Tom

#84 h Mar 17, 2025, 6:34 AM

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I will write down my solution for some reason.

Indeed, there are only three sets of solutions for this problem. Solutions are:
$(a)$ . $f(x)=kx^2$ , for all $x \in \mathbb {Z}$ , where $k$ is a constant.
$(b)$ . $f(x)=0$ for all even $x$ , and $f(x)=k$ for all odd $x$ where $k$ is a constant.
$(c)$ . $f(x)=k$ for all odd $x$ , $f(x)=0$ for all $x \equiv 0 \pmod 4$ , $f(x)=4k$ for all $x \equiv 2 \pmod 4$ .

It's easy to prove that $f(0)=0$ , and $f(x)=f(-x)$ .

Lemma (1): If $f(a)=0$ , then $f(b)=f(c)$ where $a+b+c=0$ .
Proof: $f(a)=0$ , then $f(b)^2+f(c)^2=2f(b)f(c)$ $\implies$ $(f(b)-f(c))^2=0$ and result follows.

Lemma (2): $f(2x)[f(2x)-4f(x)]=0$ , for all $x \in \mathbb{Z}$ .
Proof: Just take $(a=x, b=x, c=-2x)$ , then result follows since $f$ is even.

Lemma (3): Let's define $g(x)=\frac {f(x)}{f(1)}$ . Then $g(1)=1$ . If we have $g(a)=a^2$ and $g(b)=b^2$ , then $g(a+b)=(a+b)^2$ or $g(a+b)=(a-b)^2$ .
Proof: Just plug $(a=a, b=b, c=-(a+b))$ , then we have $[g(a+b)-(a+b)^2]*[g(a+b)-(a-b)^2]=0$ , and the result follows.

If $f(1)=0$ , then by Lemma (1) we have that $f(a+1)=f(a)$ , so our function is const, so $f(x)=0$ in this case.

Otherwise by using Lemma (2) we get that $f(2)=0$ or $f(2)=4f(1)$ .

If $f(2)=0$ , then by applying Lemma (1) we have that $f(x)=0$ for all even x, and $f(x)=k$ for all odd $x$ .

Now assume that $f(2)=4f(1)$ . Let's define $g(x)=\frac {f(x)}{f(1)}$ . Then $g(1)=1$ and $g(2)=4$ . Let $g(3)=d$ , then by taking $(a=1, b=2, c=-3)$ we will get that $(a-1)(a-9)=0$ . We will consider each cases here.

(Case 1)
Let $g(3)=1$ , Then we can get $g(4)=0$ by checking small values. Since $g(4)=4$ , by using Lemma (1), we will get $g(n+4)=g(n)$ , for all $n$ which gives us Solution(c).

(Case 3)
Let $g(3)=9$ , in this case we can easily induct on natural numbers using Lemma (3), which gives us Solution(a).