Problem
ALG-B3-M06-P008 A Binomial Correction
#8
★★★☆☆ Level 3 of 5
Find all \(f:\mathbb Z\to\mathbb Z\) such that \(f(m+n)=f(m)+f(n)+mn\).
Use \(\binom{m+n}{2}\).
Let \(B(n)=\binom{n}{2}=\frac{n(n-1)}{2}\). Then \(B(m+n)=B(m)+B(n)+mn\). Hence \(g(n)=f(n)-B(n)\) is additive on \(\mathbb Z\), so \(g(n)=cn\). The answer is \(f(n)=\binom{n}{2}+cn\), where \(c\in\mathbb Z\).
The binomial form is convenient because it is integral for all integers \(n\).