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Source: Bulgaria Spring Mathematical Competition 2025 9.4

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#1 h Mar 30, 2025, 12:39 PM

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Determine all functions $f: \mathbb{Z}_{>0} \to \mathbb{Z}_{>0}$ such that $f(a) + 2ab + 2f(b)$ divides $f(a)^2 + 4f(b)^2$ for any positive integers $a$ and $b$ .

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#2 h Mar 30, 2025, 1:43 PM • 1 Y

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Answer is $f(x)=x^2$ for all positive integers. Let $P(a,b)$ be the assertion of $f(a)+2f(b)+2ab|f(a)^2+4f(b)^2$ .
Claim: $f(1)=1$ .
Proof: $P(1,1)$ gives $3f(1)+2|5f(1)^2$ or $3f(1)+2|20$ so $f(1)=1$ or $f(1)=6$ . Suppose that $f(1)=6$ . $P(1,5)$ yields $6+2f(5)+10|36+4f(5)^2$ or $f(5)+8|2f(5)^2+18$ or $f(5)+8|146$ . Thus, $f(5)=65$ or $f(5)=138$ . However, $P(5,1)$ gives $f(5)+22|f(5)^2+144$ . If $f(5)=65$ , then $3|87|65^2+144$ and if $f(5)=138$ , then $5|160|138^2+144$ which are not correct. Hence $f(1)=1$ .
Claim: $f(a)\leq a^2$ .
Proof: $P(a,1)$ yields $2f(a)+2a+1|4f(a)^2+1$ or $2f(a)+2a+1|2(2a^2+2a+1)$ or $2f(a)+2a+1|2a^2+2a+1$ hence $f(a)\leq a^2$ .
Claim: For sufficiently large primes $p$ , we have $f(p)=p^2$ .
Proof: $P(a,a)$ implies $3f(a)+2a^2|20a^4$ . We get $3f(p)+2p^2|20p^4$ and if $(p,f(p))=1$ , then $3f(p)+2p^2|20$ which is impossible. If $f(p)=pk$ such that $(p,k)=1$ , then $3k+2p|20p^3$ or $3k+2p|20$ which is impossible again. Thus, $p^2|f(p)$ and since $f(p)\leq p^2$ , we have $f(p)=p^2$ for sufficiently large primes.
Claim: $f(a)=a^2$ for all positive integers.
Proof: Pick $p>2024!a$ . By $P(a,p)$ , we see that $f(a)+2pa+2p^2|f(a)^2+4p^4$ or $f(a)+2pa+2p^2|4p^2(p^2+(p+a)^2)=4p^2(2p^2+2pa+a^2)$ and since $(f(a)+2pa+2p^2,p)=1$ , we have $f(a)+2pa+2p^2|4(2p^2+2pa+a^2)$ or $f(a)+2pa+2p^2|4(a^2-f(a))$ . Since $f(a)+2pa+2p^2$ tends to positive infinity while right hand side is constant, $f(a)=a^2$ as desired. $\blacksquare$

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bo18

#3 h Mar 30, 2025, 5:02 PM

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@above:
Just like my solution on competition
even the transformations on P(a, p)
Bravo

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