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Source: 2024 JMO preliminary p8

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#1 h Jan 9, 2024, 7:36 AM

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Function $f:\mathbb{Z}_{\geq 0}\rightarrow\mathbb{Z}$ satisfies
$f(m+n)^2=f(m|f(n)|)+f(n^2)$ for any non-negative integers $m$ and $n$ . Determine the number of possible sets of integers $\{f(0), f(1), \dots, f(2024)\}$ .

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#3 h Jan 10, 2024, 5:37 AM

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answer

:maybe:

sol

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#5 h May 2, 2024, 12:46 AM

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The answer is $\boxed{992}$ . Let $P(m,n)$ be the given assertion. Notice $P(0,0)$ gives $f(0)^2=2f(0)$ so either $f(0)=0$ or $f(0)=2$ .

Case 1: $f(0)=0$
The assertion $P(m,0)$ gives $f(m)^2=0\Rightarrow f(m)=0$ . Which can be checked to work.

Case 2: $f(0)=2$
The assertion $P(0,n)$ gives $f(n)^2=2+f(n^2)$ . The assertion $P(m,0)$ gives $f(m)^2=f(2m)+2$ . Combined we get $f(n^2)=f(2n)$ . The assertion $P(0,1)$ gives $f(1)^2=2+f(1)$ so either $f(1)=2$ or $f(1)=-1$ . If $f(1)=-1$ then $f(2)=-1$ and $P(1,1)$ gives a contradiction. Now assume $f(1)=2\Rightarrow f(2)=2$ . We get from $P(m,1)$ that $f(m+1)^2=f(2m)+2\Rightarrow f(2m+2)=f(2m)\Rightarrow$ $f(2m)=2\Rightarrow f(m^2)=2$ . Thus we get $f(n)^2=4\Rightarrow f(n)=\pm 2$ . Thus $f(x)=2$ whenever $x$ is even or a square and $f(x)$ can be either $2$ or $-2$ when $x$ is not even or a square. It is easy to check that all such functions work.

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#8 h Oct 13, 2024, 8:01 AM

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I think the correct answer should be 991, because the number of possible sets in Case 2 should be 990. It seems like you may have made a small counting mistake. The number of integers from 3 to 2024 that are neither even nor perfect squares is 990.

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#9 h Apr 30, 2025, 3:54 AM

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$P(0, 0)\Longrightarrow f(0)=0, 2$
if $f(0)=0$ , then $P(m, 0)\Longrightarrow f(m)^2=0\Longrightarrow f(m)=0$

Let $f(0)=2$ from now on
$P(0, 1)\Longrightarrow f(1)^2=2+f(1)\Longrightarrow f(1)=-1, 2$
if $f(1)=-1$ then $P(1, 1)\Longrightarrow f(2)^2=-2$ , contradiction, so $f(1)=2$
$P(m, 1)\Longrightarrow f(m+1)^2=f(2m)+2$ , $P(m, 0)\Longrightarrow f(m)^2=f(2m)+2$ , so $f(m+1)=\pm f(m)$ , and so $f(m)=\pm2$
and so the original FE is now $4=f(2m)+f(n^2)$ , which means even numbers and perfect squares must have $f(a)=2$ , the others can have $f(a)=2, -2$

The number of sets would be just $2$ as sets don't count duplicates. If it is multiset then the answer is $991$ .

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