Problem
ALG-B3-M06-P018 Arguments of the Same Parity
#18
★★★★★ Level 5 of 5
Find all \(f:\mathbb Z\to\mathbb Z\) such that \(f(m+n)-f(m-n)=4mn\) for all \(m,n\).
Put \(u=m+n\), \(v=m-n\). Then \(u\) and \(v\) have the same parity.
Notice that \(4mn=(m+n)^2-(m-n)^2\). Let \(u=m+n\), \(v=m-n\). Then \(u\) and \(v\) have the same parity, and the condition says \(f(u)-u^2=f(v)-v^2\). Hence \(f(n)-n^2\) is constant separately on even and on odd \(n\). Thus all solutions are \(f(n)=n^2+c_0\) for even \(n\) and \(f(n)=n^2+c_1\) for odd \(n\), where \(c_0,c_1\in\mathbb Z\). The check is immediate because \(m+n\) and \(m-n\) always have the same parity.
Key trap: the equation does not connect the two parity classes.