Natural function and cubelike expression

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Source: Philippine Mathematical Olympiad 2025 P8

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#1 h Mar 9, 2025, 3:42 PM

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Let $\mathbb{N}$ be the set of positive integers. Find all functions $f : \mathbb{N} \to \mathbb{N}$ such that for all $m, n \in \mathbb{N}$ , $m^2f(m) + n^2f(n) + 3mn(m + n)$ is a perfect cube.

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#2 h Mar 16, 2025, 3:22 PM • 1 Y

Y by shafikbara48593762

For any prime number $p$ , the $m^2f(m) + p(pf(p) + 3m(m+p))$ is a perfect cube. This means that for fixed $m$ , the polynomial $X^3 - m^2f(m)$ has a root in $\mathbb{F}_p$ for any prime $p$ . It is known that this implies that $m^2f(m)$ is a perfect cube. Let $m^2f(m) = g(m)^3$ for any positive integer $m$ . For any prime $p$ , $p \mid g(p)$ , so $g(p) \ge p$ .

Now, since $g(p)^3 + 3p^2 + 3p + f(1)$ is a perfect cube for all prime $p$ , we have that $3g(p)^2 + 3g(p) + 1 = (g(p) + 1)^3 - g(p)^3 \le 3p^2 + 3p + f(1)$ When $p$ is large enough, this can only hold when $g(p) \le p$ . This implies that $g(p) = p$ for all sufficiently large prime $p$ .

Fix $m$ , and plug in $n = p$ , so that $m^2(f(m) - m) + (m+p)^3$ is a perfect cube for any prime $p$ . When $p$ is large enough, the gap between consecutive cubes becomes larger than $|m^2(f(m) - m)|$ , which implies that $f(m) = m$ . Therefore, the only possible solution is $f(m) = m$ for all positive integer $m$ , which obviously satisfies the condition.

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#3 h Mar 18, 2025, 8:19 AM

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@above Can you please explain the well known lemma you use?

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