Nordic 2025 P1

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#1 h Tuesday at 12:32 PM

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Let $n$ be a positive integer greater than $2$ . Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfying:
$(f(x+y))^{n} = f(x^{n})+f(y^{n}),$ for all integers $x,y$

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#2 h Tuesday at 12:46 PM

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anirbanbz wrote:

Let $n$ be a positive integer greater than $2$ . Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfying:
$(f(x+y))^{n} = f(x^{n})+f(y^{n}),$ for all integers $x,y$

Let $P(x,y)$ be the assertion $f(x+y)^n=f(x^n)+f(y^n)$

$P(0,0)$ $\implies$ $f(0)^n=2f(0)$ and so $f(0)=0$
$P(x,0)$ $\implies$ $f(x)^n=f(x^n)$ and so $P(x,y)$ may be written $f(x+y)^n=f(x)^n+f(y)^n$

And so $f(2x)^n=2f(x)^n$ and so $\boxed{f(x)=0\quad\forall x\in\mathbb Z}$ , which indeed fits

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#3 h Tuesday at 3:13 PM

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Solution

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#4 h Tuesday at 3:21 PM • 2 Y

Y by amapstob, observer04

ffmfmemrmamts hlast hteorm.

Take $x = y = 0$ to get $f(0)^n = 2f(0)$ so $f(0) = 0$ . Then take $y = 0$ to get $f(x)^n = f(x^n)$ , so this becomes $f(x + y)^n = f(x)^n + f(y)^n$ . Then note that by Fermat's Last Theorem, this means that at least one of $f(x), f(y)$ , and $f(x + y)$ is $0$ . Thus, all but at most one value of $f$ gets mapped to zero, and if that value $v$ exists, then taking $x = v$ gives a contradiction when $x+y, y \ne v$ .

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#5 h Tuesday at 3:34 PM • 1 Y

Y by YaoAOPS

YaoAOPS wrote:

ffmfmemrmamts hlast hteorm.

Take $x = y = 0$ to get $f(0)^n = 2f(0)$ so $f(0) = 0$ . Then take $y = 0$ to get $f(x)^n = f(x^n)$ , so this becomes $f(x + y)^n = f(x)^n + f(y)^n$ . Then note that by Fermat's Last Theorem, this means that at least one of $f(x), f(y)$ , and $f(x + y)$ is $0$ . Thus, all but at most one value of $f$ gets mapped to zero, and if that value $v$ exists, then taking $x = v$ gives a contradiction when $x+y, y \ne v$ .

I was gonna try to find a solution using this but decided not to, its a really fun theorem to use in olympiads though :play_ball:

:play_ball:

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