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Source: Own, PDC002-P7

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#1 h Apr 21, 2025, 11:25 AM

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Let $f : \mathbb{Z}^+ \to \mathbb{Z}^+$ be a function such that
$f(m) + mn + n^2 \mid f(m)^2 + m^2 f(n) + f(n)^2$ for all $m, n \in \mathbb{Z}^+$ . Find all such functions $f$ .

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#2 h Apr 21, 2025, 1:27 PM

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Denote the given assertion by $P\left(m,n\right)$ .
$P\left(n,n\right)$ : $f\left(n\right)+2n^2\mid 2f\left(n\right)^2+n^2f\left(n\right)\equiv 6n^4$ Denote this by $Q\left(n\right)$ .
$Q\left(1\right)$ :
$f\left(1\right)+2\mid6\to f\left(1\right)=1\vee f\left(1\right)=4$ If $f\left(1\right)=4$ we get from $P\left(1,2\right)$ :
$10\mid f\left(2\right)^2+f\left(2\right)+16$ In particular:
$5\mid f\left(2\right)^2+f\left(2\right)+1\mid \left(2f\left(2\right)+1\right)^2+3$ Which is impossible, thus $f\left(1\right)=1$ .
$P\left(m,1\right)$ :
$f\left(m\right)+m+1\mid f\left(m\right)+m^2+1\equiv 2\left(m^2+m+1\right)$ Suppose for some m we have equality, from $Q\left(m\right)$ :
$4m^2+m+1\mid6m^4\overset{\gcd\left(4m^2+m+1,m\right)=1}{\to}4m^2+m+1\mid6\to m=1$ Which is false since $f\left(1\right)\ne4$ .
This means:
$f\left(m\right)+m+1\leq\frac{2\left(m^2+m+1\right)}{2}=m^2+m+1\to f\left(m\right)\leq m^2$ Now let $p>100$ be prime, notice from $Q\left(p\right)$ we have a divisor of $6p^4$ , namely $f\left(p\right)+2p^2$ , which is in $(2p^2,3p^2]$ , thus it must be $3p^2$ thus $f\left(p\right)=p^2$ for large primes.
Now $P\left(p,n\right)$ for large primes p gives:
$p^2+pn+n^2\mid f\left(n\right)^2+p^2f\left(n\right)+p^4$ Note also
$p^2+pn+n^2\mid n^4+p^2n^2+p^4$ thus subtracting gives:
$p^2+pn+n^2\mid \left(n^2-f\left(n\right)\right)\left(n^2+f\left(n\right)+p^2\right)$ Now assume by contradiction that $n^2-f\left(n\right)>0$ for some n and denote this value by k, we claim p can be chosen so that $\gcd\left(k,p^2+pn+n^2\right)=1$ .
Lemma: For every prime q and integer $n$ there exists an integer $q\nmid r$ s.t.
$q\nmid r^2+rn+n^2$ Proof: Suppose otherwise, we have from $r=\pm1$ :
$q\mid n^2\pm n +1\to q\mid 2n$ If $q\mid n$ then $n^2+rn+r^2\equiv_q r^2$ and $r=1$ gives contradiciton.
If $q\nmid n$ then $q=2$ and then $r^2+rn+n^2\equiv_2 r^2+r+1$ and $r=1$ still gives contradiction.
Using the lemma, for every prime $q\mid k$ we choose $p\equiv r \mod{q}$ such that $q\nmid p^2+pr+r^2$ . This is equivalent to choosing p to be some $a\mod{b}$ for some coprime a and b from the chinese remainder theorem, then infinitely many such p exists from dirichlet's theorem giving us:
$p^2+pn+n^2\mid k\left(n^2+f\left(n\right)+p^2\right)\to p^2+pn+n^2\mid n^2+f\left(n\right)+p^2\to pn\leq f\left(n\right)$ And choosing sufficiently large p concludes, thus $\forall n\in\mathbb{Z}^+:f\left(n\right)=n^2$ which fits.

This post has been edited 1 time. Last edited by rrrMath, Apr 21, 2025, 1:30 PM

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