IMO Shortlist 2013, Number Theory #1

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127 posts

#1 h Jul 10, 2014, 6:08 AM • 27 Y

Y by Davi-8191, abab, Muaaz.SY, itslumi, mathematicsy, megarnie, rachelshi, jhu08, justJen, GeoMetrix, fluff_E, son7, Lamboreghini, Stuart111, hectorleo123, Adventure10, Mango247, trigadd123, ItsBesi, Akacool, rightways, pomodor_ap, cubres, and 4 other users

Let $\mathbb{Z} _{>0}$ be the set of positive integers. Find all functions $f: \mathbb{Z} _{>0}\rightarrow \mathbb{Z} _{>0}$ such that
$m^2 + f(n) \mid mf(m) +n$
for all positive integers $m$ and $n$ .

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nicetry007

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nicetry007

#2 h Jul 10, 2014, 7:14 AM • 57 Y

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Setting $m = f(n)$ , we get $f(n)^2 + f(n) \mid f(n)f(f(n)) + n \Rightarrow f(n) \mid n \Rightarrow f(n) \leq n \; \forall \; n \in \mathbb{N}$ .
This implies $f(1) = 1$ .
Setting $m=n$ , we get $n^2 + f(n) \mid nf(n) + n \Rightarrow n^2+f(n) \leq nf(n) + n \Rightarrow n^2 - n \leq (n-1)f(n)$
This implies $n \leq f(n) \;\forall \; n \geq 2 \Rightarrow f(n) = n \; \forall n \in \mathbb{N}$ .

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AkosS

113 posts

AkosS

#3 h Jul 10, 2014, 7:33 AM • 11 Y

Y by Muaaz.SY, rachelshi, jhu08, megarnie, son7, Adventure10, Mango247, cubres, and 3 other users

Let $m=n$ , then
$m^2+f(m)|m(f(m)+1)\Rightarrow f(m)=\frac{k_m m^2-m}{m-k_m}$
for some $k_m\in\mathbb{Z}^+$ , so $1<k_m<m$ and thus $f(m)\leq m^3-m^2-m$ , as with $k$ growing the numerator grows and the denominator shrinks, for $m=2$ , this gives us $k=1$ and $f(2)=2$ .

Now use $(2,m)$ , then for some $l_m\in\mathbb{Z}^+$
$4+f(m)|4+m\Rightarrow f(m)=\frac{m+4}{l_m}-4\leq m$
Using $m=1$ and $m=3$ in $4+f(m)|4+m$ we get $f(1)=1$ and $f(3)=3$ , so we can assume $m\geq 4$
Now look at the following:
$\frac{m(mk_m-1)}{m-k_m}=f(m)\leq m$
Use $k_m=2$ , we get:
$\frac{m(2m-1)}{m-2}-m=\frac{m(m+1)}{2}>0$
This means that only $k_m=1$ can hold, meaning $f(m)=m$ .

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alibez

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alibez

#4 h Jul 10, 2014, 11:48 AM • 10 Y

Y by rachelshi, jhu08, megarnie, son7, kamatadu, Adventure10, cubres, and 3 other users

lyukhson wrote:

Let $\mathbb{Z} _{>0}$ be the set of positive integers. Find all functions $f: \mathbb{Z} _{>0}\rightarrow \mathbb{Z} _{>0}$ such that
$m^2 + f(n) \mid mf(m) +n$
for all positive integers $m$ and $n$ .

we know : $f(p)^{2}+f(p)\mid f(p)f(f(p))+p\Rightarrow f(p)=1\: or\: p$ ( $p$ is prime number)

$i)$ if we have Infinitely many prime numbers such that $f(p)=p$ so we have:

$m^{2}+p\mid mf(m)+p\Rightarrow m^{2}+p\mid mf(m)-m^{2}\Rightarrow mf(m)-m^{2}=0\rightarrow f(m)=m$

$ii)$ if we don't have Infinitely many prime numbers such that $f(p)=p$ !

$m=n=p\rightarrow p^{2}+1\mid 2p$ and it is not true for Infinitely many prime numbers . so $f(m)=m$ is unique solution.

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gobathegreat

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gobathegreat

#5 h Jul 10, 2014, 1:21 PM • 8 Y

Y by rachelshi, jhu08, tony88, Adventure10, cubres, and 3 other users

lyukhson wrote:

Let $\mathbb{Z} _{>0}$ be the set of positive integers. Find all functions $f: \mathbb{Z} _{>0}\rightarrow \mathbb{Z} _{>0}$ such that
$m^2 + f(n) \mid mf(m) +n$
for all positive integers $m$ and $n$ .

Let $P(m,n)$ be assertion of $m^2 + f(n) \mid mf(m) +n$
$P(f(n),n)$ implies $f(n)$ divides $n$ so $f(n) \leq n$ .This implies $f(1)=1$ . Now suppose for some $n$ we have $f(n)=n$ (this is possible since $f(1)=1$ ). We will prove that $f(n+1)=n+1$ .
Consider:
$P(n,n+1)$ :
Then $n^2 + f(n+1) \mid nf(n) +n+1$ from which we obtain $f(n+1) \leq n+1$
$P(n+1,n)$ :
Then $(n+1)^2 + n \mid (n+1)f(n+1) +n$ from which we obtain $n+1 \leq f(n+1)$ .Together this implies $f(n+1) \leq n+1 \leq f(n+1)$ so $f(n+1)=n+1$ .
Since $f(1)=1$ we see that $f(n)=n$ for every positive integer $n$ .

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mathuz

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mathuz

#6 h Jul 10, 2014, 10:33 PM • 8 Y

Y by Mathuzb, rachelshi, jhu08, Adventure10, Mango247, cubres, and 2 other users

It's inequality problem. We have that $f(n)\ge n$ and there exists $k\in N$ such that $f(k)\le n+f(1)-1<2k$ . So $f(k)=k$ . So for all $n\in N$ $f(n)=n.$

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fclvbfm934

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fclvbfm934

#7 h Jul 11, 2014, 6:57 PM • 7 Y

Y by rachelshi, jhu08, Adventure10, Mango247, cubres, and 2 other users

Solution with thkim1011:
Let $m = n$ and we get $m^2 + f(m) | mf(m) + m$ so we have
$m^2 + f(m) \le mf(m) + m$
which rearranges to $f(m) \ge m$ for all $m \ge 2$ . Now let $m = f(n)$ and we get
$f(n)^2 + f(n) | f(f(n))\cdot f(n) + n$
so $f(n) |n$ but $f(n) \ge n$ , so the only viable possibility is $f(n) = n$ for $n \ge 2$ . For $n = 1$ , we have $f(1) | 1$ , so $f(1) = 1$ . It is easy to check that $f(n) = n$ satisfies the condition.

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Number1

355 posts

Number1

#8 h Jul 12, 2014, 10:23 AM • 8 Y

Y by MiMi1376, randomasdf97, rachelshi, Adventure10, cubres, and 3 other users

If we put $m=n$ we get $f(m)\geq m$ for all $m \geq 2$ (*).

If we put $m=n=2$ we get $4+f(2)|2f(2)+2$ and thus $f(2) =2$ .

If we put $m=2$ we get $4+f(n)|4+n$ and thus $f(n)\leq n$ for all $n$ (**).

If we use (*) and (**) we get $f(n)=n$ for all $n$ .

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MiMi1376

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MiMi1376

#9 h Jul 12, 2014, 11:21 AM • 6 Y

Y by rachelshi, Adventure10, Mango247, cubres, and 2 other users

it was really easy!!!
this problem was proposed by Mohsen Jamaali

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math_genie

40 posts

math_genie

#10 h Jul 14, 2014, 4:21 PM • 6 Y

Y by rachelshi, Adventure10, Mango247, cubres, and 2 other users

Nice solution everyone. Was quite easy

My first impression was it was very similar to IMO 2004 shortlist problem (N3). Indeed, the solution was very similar.

First, I noted that $f(n) = n$ was a solution (as almost always).
And I noticed that we need to just prove either $f(1) = 1$ or $f(p) = p$ for all $p$ prime.

(*)

[Taking $m = n = p$ gives us
$p^2 + f(p) \mid p(f(p) + 1)$
Since $p^2 + f(p)$ does not divide $f(p) + 1$ (as its bigger) thus,
$p \mid p^2 + f(p)$ i.e. $p \mid f(p)$
Therefore $f(p) \geq p$

If $f(1) = 1$ then take $m = 1$ , $n = p$ and we get $1 + f(p) \mid p + 1$ . Thus $f(p) = p$ for all p prime. (take inequality)
And, if we show $f(p) = p$ , then taking m = p we get
$p^2 + f(n) \mid p^2 + n$
Thus $p^2 + f(n) \mid (n - f(n))$ .
Choosing $p$ large enough, we get the quotient tends to 0. (as $n - f(n)$ is fixed but $p$ can be as large as you want).
Therefore, $f(n) = n$ for all $n$ in naturals]

To prove (*) - it didn't hit me to take $m = f(n)$ as nicetry007 did. (You could probably finish the proof like that)

Instead, what I did was the following:
Let $p$ be a prime and take $n = p$ .
Let $q$ be a prime dividing $f(p)$ , and take $m = q$ . We get:

$q^2 + f(p) \mid qf(q) + p$
But $q\mid q^2 + f(p)$ !
Thus, we get $q \mid p$ , or $p = q$ .
So, we have $f(p) = p^i$ , where $i$ depends on $p$ .

Now, take $m = n = p$ .
We get
$p^2 + p^i \mid p^{i+1} + p$ .

(1)

Case 1 : $i \geq 3$
Easy contradiction to (1) as $p^2 | LHS$ but not the $RHS$ .

Case 2: $i = 2$
Then we get $2p^2 \mid p^3 + p$
or, $2p \mid p^2 + 1$ , this means $p\mid1$ . Contradiction.

Thus, $f(p) = p$ and we're done.

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AnonymousBunny

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AnonymousBunny

#11 h Jul 14, 2014, 5:51 PM • 5 Y

Y by rachelshi, Adventure10, Mango247, cubres, and 1 other user

Plugging $m=n,$
$m^2 + f(m) \leq mf(m) + m \implies m^2 + f(m) \leq mf(m) + m \implies m \leq f(m).$
Plugging $n=f(m),$
$f(m) \mid f(m)^2 + f(m) \mid mf(f(m)) + f(m) \implies f(m) \mid m \implies f(m) \leq m.$
Hence, $\boxed{f(m)=m \quad \forall \ m \in \mathbb{N}}.$

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numbertheorist17

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numbertheorist17

#12 h Aug 5, 2014, 2:41 PM • 5 Y

Y by girishvar12, rachelshi, Adventure10, cubres, and 1 other user

Solution

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codyj

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codyj

#13 h Aug 6, 2014, 8:05 AM • 7 Y

Y by rachelshi, Adventure10, Mango247, cubres, and 3 other users

Let $m=n$ to get $n^2+f(n)\mid nf(n)+n$ , so $n^2+f(n)\le nf(n)+n$ , or $f(n)\ge n$ . In the case of $n=2$ , $4+f(2)\mid 2f(2)+2$ , and $\frac{2f(2)+2}{f(2)+4}=2-\frac6{f(2)+4}\in\mathbb{Z}$ . Therefore, $f(2)+4\in\{-6,-3,-2,-1,1,2,3,6\}$ , and the only solution with a positive $f(2)$ is $f(2)=2$ . Let $m=2$ to get $2^2+f(n)\mid2f(2)+n$ , so $4+f(n)\le2f(2)+n=4+n$ , or $f(n)\le n$ . Therefore, $f(n)=n$ is the only solution, and we can easily verify that $m^2+n\mid m^2+n$ .

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smy2012

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smy2012

#14 h Aug 7, 2014, 4:06 AM • 3 Y

Y by Adventure10, Mango247, cubres

Very easy problem.

I find 2 solutions (the same as #8 and #11) at once.

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Dexenberg

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Dexenberg

#15 h Dec 14, 2014, 4:50 PM • 6 Y

Y by rachelshi, Adventure10, Mango247, cubres, and 2 other users

$m=2;n=1 \Rightarrow 4+f(1) | 2f(2)+1$
$m=1;n=2 \Rightarrow 1+f(2)|f(1)+2 \Rightarrow f(1)+1 \geq f(2) \Rightarrow$
$4+f(1) \geq \dfrac{2f(2)+1}{2} \Rightarrow 4+f(1)=2f(2)+1$ , yielding to $1+f(2) | 2f(2)-1 \Rightarrow f(2)=2;f(1)=1$
$f(1)=1 -> 1+f(n) | n+1 \Rightarrow n \geq f(n); m^2+1 | mf(m)+1 \Rightarrow m \leq f(m)$ , and now we have the conclusion: $f(x)=x$

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RedFireTruck

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RedFireTruck

#147 h Aug 1, 2024, 5:48 AM • 1 Y

Y by cubres

Letting $m=f(n)$ gives $(f(n)^2+f(n))|(f(n)f(f(n))+n)$ so $f(n)|n$ .

Since $f(1)|1$ , $f(1)=1$ .

Plugging in $n=1$ gives $(m^2+1)|(mf(m)+1)$ . Consider $x\ge 2$ . Since $f(x)<x$ gives $(x^2+1)|(xf(x)+1)<(x^2+1)$ but $f(x)|x$ , we must have $\boxed{f(x)=x}$ for all $x$ .

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ezpotd

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ezpotd

#149 h Sep 8, 2024, 1:22 AM • 1 Y

Y by cubres

I claim the answer is only the identity, it is easy to verify that this works.

Take $m = f(n)$ , we have $f(n) \mid f(n)^2 + f(n) \mid f(n)f(f(n)) + n$ so $f(n) \mid n$ , forcing $f(1) = 1$ . Then set $n = 1$ , we have $f(1) - 1 \le m (f(m) - m)$ , implying $m \ge f(m)$ , since we already know $f(m) \mid m$ we have $m = f(m)$ for all $m$ as desired.

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AshAuktober

987 posts

AshAuktober

#150 h Sep 24, 2024, 6:11 AM • 1 Y

Y by cubres

Denote the given assertion by $P(m, n)$ .
We claim the full solution set is $f(x) = x$ for all $x$ . This function works, so we now prove it is the only such one.

Claim: $f(x) \ge x$ for all $x$ .

Proof: Observe that $P(m, 1) \implies m^2 + f(1) \mid mf(m) + 1$ $\implies mf(m) + 1 \ge m^2 + f(1) \ge m^2 + 1$ $\implies f(m) \ge m.$ $\square$

Claim: $f$ has a fixed point (in particular, $f(2) = 2$ ).

Proof: Note that $P(2, 2) \implies 4 + f(2) \mid 2f(2) + 2$ $\implies 4 + f(2) \mid 6$ $\implies 4 + f(2) \le 6 \implies f(2) \le 2.$ But from the first claim, $f(2) \ge 2$ . Therefore we indeed have $f(2) = 2$ . $\square$

Claim: $f$ is the identity function.

Proof: We have that $P(2, m) \implies 4 + f(n) \mid 4 + n$ $\implies 4 + f(n) \le 4 + n \implies f(n) \le n.$ But from the first claim $f(n) \ge n$ . So we have $f(n) = n$ for all $n$ . $\square$

This final claim was what we wanted to prove, QED.

This post has been edited 1 time. Last edited by AshAuktober, Sep 24, 2024, 6:13 AM

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pie854

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pie854

#151 h Nov 7, 2024, 8:14 PM • 1 Y

Y by cubres

Putting $m=n$ it becomes obvious that $f(m)\geq m$ , for any $m$ . So, let's define $g:\mathbb Z_{>0} \to \mathbb Z_{\geq 0}$ such that $g(n)=f(n)-n$ . Then, $m^2+g(n)+n\mid m^2+mg(m)+n \implies m^2+g(n)+n \mid mg(m)-g(n) \qquad (1)$ Note that if $g(k)\leq g(1)$ for some $k$ then $k$ must be bounded. So there is a $N$ such that $g(n)\geq g(1)$ for $n\geq N$ . Taking $n\geq N$ and putting $m=1$ in $(1)$ we get $g(n)+n+1\mid g(n)-g(1).$ This forces $g(n)=g(1)$ . Now taking $m=n\geq N$ in $(1)$ we get $n^2+n+g(1)\mid g(1)(n-1).$ For large enough $n$ this forces $g(1)=0$ . Thus it follows that $g(n)=0$ for all $n$ . Hence, $f(n)=n$ for all $n$ which clearly works.

This post has been edited 3 times. Last edited by pie854, Nov 7, 2024, 8:16 PM

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Assassino9931

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Assassino9931

#152 h Nov 16, 2024, 5:38 AM • 1 Y

Y by cubres

I thought I posted this before, but maybe not: here is a generalization from Bulgaria's Balkan MO 2014 TST - for a fixed positive integer $k$ , consider $m^2 + f(n^k) \mid mf(m) + n^k$ .

To solve this, note that $m=f(n^k)$ yields that $f(n^k)$ divides $n^k$ , in particular $f(1) = 1$ . But then $m^2 + 1 \mid mf(m) + 1$ , so $f(m) \geq m$ for all $m$ , thus $f(n^k) = n^k$ for all $n$ . Back to the initial condition, we want $m^2 + n^k \mid mf(m) + n^k$ , i.e. $m^2 + n^k \mid m(f(m) - m)$ . Fixing $m$ , the divisor attains infinitely many values as $n$ varies, thus $m(f(m) - m) = 0$ for all $m$ , implying $f(m) = m$ for all $m$ , which works.

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Eka01

204 posts

Eka01

#153 h Dec 5, 2024, 8:08 AM • 1 Y

Y by cubres

Put $m=f(n)$ to get $f(n) | n$ . Let $f(m)=\frac{m}{d} ;d|m$ and Now $m=n$ to get $m^2 +\frac{m}{d} | \frac{m^2}{d}+m^2$ .
This implies $dm^2+m |m^2+dm^2$ or $dm+1 |m+d$ . Now $dm+1 \leq m+d$ so $d=1$ or $m=1$ so $d=1$ in the latter case as well. Hence $f(m)=m$ .

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lelouchvigeo

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lelouchvigeo

#154 h Dec 9, 2024, 10:18 AM • 1 Y

Y by cubres

Sol

This post has been edited 1 time. Last edited by lelouchvigeo, Dec 9, 2024, 10:18 AM
Reason: ..

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smileapple

1010 posts

smileapple

#155 h Dec 28, 2024, 7:53 AM • 1 Y

Y by cubres

Let $P(m,n)$ denote the given relation. Setting $P(2,2)$ yields that $(f(2)+4)\mid(2f(2)+2)$ , so that $(f(2)+4)\mid6$ and thus $f(2)=2$ .

Now setting $P(2,n)$ yields that $(f(n)+4)\mid(n+4)$ , while setting $P(n,2)$ yields that $(n^2+2)\mid(nf(n)+2)$ . Since $n$ and $f(n)$ are always positive integers, it follows that $f(n)+4\le n+4$ and $n^2+2\le nf(n)+2$ , or equivalently $f(n)=n$ , for all integers $n$ . Hence $f$ is the identity. $\blacksquare$

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TestX01

338 posts

TestX01

#156 h Jan 8, 2025, 4:46 AM • 1 Y

Y by cubres

uwu~ i think this is different from other ppl

Let $P(m,n)$ be the assertion given.

$P(1,n)$ , yields $1+f(n)\mid f(1)+n$ . Let $n=p-f(1)$ for some prime $p$ large enough. Then, $1+f(p-f(1))\mid p$ , but this means (as $f\geq 1$ by codomain), $1+f(p-f(1))=p$ , or $f(p-f(1))=p-1$ . Now, $P(m,p-f(1))$ gives $m^2+p-1\mid mf(m)+p-f(1)$ . Hence, $m^2+p-1\mid mf(m)+p-f(1)-m^2-p+1=mf(m)-m^2+1-f(1)$ .

The RHS is constant for $m$ constant, but the LHS tends to infinity as $p\to \infty$ (there is an infinitude of primes). This is impossible unless the RHS is $0$ , or $mf(m)-m^2=f(1)-1$ . Factoring, and noting $f(1)\geq 1$ , $m(f(m)-m)\geq 0$ , hence $f(m)\geq m$ . Now, suppose that $f(x)\neq x$ for infinitely many positive integers $x$ . Then, there must be arbitrarily large $m$ such that $f(m)>m$ , (as equality won't hold) thus by discrete inequality, $f(m)-m\geq 1$ so $f(1)-1=m(f(m)-m)\geq m$ which is a contradiction as the LHS is bounded.

Now, this implies that for some constant $C$ , $f(x)=x \quad \forall x>C$ , now for some $m>C$ , considering $P(m,n)$ gives $m^2+f(n)\mid m^2+n$ . As the RHS is non-negative, $m^2+f(n)\leq m^2+n$ , or $f(n)\leq n$ for all positive integers $n$ . Combining with the previous bound of $f(m)\geq m$ , we conclude that $f(x)=x$ for all positive integers $x$ .

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Maximilian113

539 posts

Maximilian113

#157 h Jan 10, 2025, 7:00 PM • 1 Y

Y by cubres

I think I have a pretty unique and short solution.

Let $n=m,$ then $m^2+f(m) \leq mf(m)+m \implies (m-f(m))(m-1) \leq 0 \implies f(m) \geq m.$

Now suppose that $n$ is such that $m|f(n).$ Then $m | m^2+f(n) | mf(m)+n \implies m | n.$ Thus letting $m=f(n)$ yields $f(n) | n \implies f(n) \leq n.$

Thus, $f(m)=m$ for all $m,$ and this is the only solution.

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math004

23 posts

math004

#158 h Feb 4, 2025, 12:02 AM • 1 Y

Y by cubres

$P(f(n),n) : f(n)^2+f(n)\mid f(n)f((n))+n\implies f(n)\mid n \implies f(n)\leq n$
$P(n,1) : n^2+1\leq n^2+f(1) \mid nf(n)+1 \implies nf(n)+1\geq n^2+1\implies f(n)\geq n.$

Whence $f\equiv n$ which Convsersely satisfies the equation.

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math-olympiad-clown

19 posts

math-olympiad-clown

#159 h Mar 3, 2025, 2:38 PM • 1 Y

Y by cubres

m=n=2 plug in : 4+f(2)|2f(2)+2 if 2f(2)+2=k*[4+f(2)] and kis greater than 1 it's easy to see it's not possible.
so k=1 gives f(2)=2.

m=2 plug in :4+f(n)|4+n this gives f(n)<=n.
n=2 plug in : m^2+2|mf(m)+2 this gives f(m)>=m.

then we have f(m)=m is the only solution :-D

.

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SSS_123

20 posts

SSS_123

#160 h Apr 12, 2025, 5:16 PM • 1 Y

Y by cubres

too big solution lol

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MuradSafarli

85 posts

MuradSafarli

#161 h Apr 13, 2025, 9:41 AM

Y by

We find from the substitution $P(2,2)$ that $f(2) = 2$ .
From $P(2,3)$ , we obtain $f(3) = 3$ .
Finally, from $P(3,m)$ , we get $m \geq f(m)$ ,
and from $P(m,2)$ , we get $f(m) \geq m$ .
Therefore, $f(m) = m$ for all natural numbers $m$ .

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InterLoop

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InterLoop

#162 h Yesterday at 12:32 AM

Y by

another ntfe
solution

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