Problem 3

Find all functions

g:\mathbb{Z}_{>0}\to\mathbb{Z}_{>0}

such that

\left(g(m)+n\right)\left(g(n)+m\right)

is a perfect square for all

m,n\in\mathbb{Z}_{>0}

Step 4 of 6: Deduce unit consecutive differences and injectivity

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g(k+1)-g(k)\in\{-1,1\},\qquad g(a)=g(b)\Longrightarrow a=b

Detailed analysis

If $g(k)=g(k+1)$ , the claim applied modulo any prime $p$ would give $k\equiv k+1\pmod p$ , impossible. If $|g(k+1)-g(k)|>1$ , choose a prime $p$ dividing this difference; then $g(k)\equiv g(k+1)\pmod p$ , and the claim again gives the same contradiction. Hence $|g(k+1)-g(k)|=1$ . More generally, if $g(a)=g(b)$ with $a\ne b$ , the rigidity claim applies for every prime $p$ , so $a\equiv b\pmod p$ for every prime. Choosing a prime larger than $|a-b|$ forces $a=b$ , so $g$ is injective.