Problem
ALG-B1-M08-P020 A quadratic equation on integers
#20
★★★★☆ Level 4 of 5
A function \(f:\mathbb Z\to\mathbb Z\) satisfies \(f(m+n)+f(m-n)=2f(m)+2f(n)\), \(f(0)=0\), \(f(1)=1\). Prove that \(f(n)=n^2\) for all \(n\in\mathbb Z\).
Substitute \(m=n\), then \(m=n\), \(n=1\), to get a recurrence.
At \(m=0\), we get \(f(n)+f(-n)=2f(n)\), so \(f(-n)=f(n)\). At \(n=1\): \(f(m+1)+f(m-1)=2f(m)+2\).
This recurrence with \(f(0)=0\), \(f(1)=1\) determines all values. We prove by induction that \(f(k)=k^2\) for \(k\ge0\). It is true for \(0,1\). If it is true for \(k\) and \(k-1\), then \(f(k+1)=2f(k)+2-f(k-1)=2k^2+2-(k-1)^2=(k+1)^2\). For negative \(n\), use \(f(-n)=f(n)\).
This is an introductory version of a quadratic functional equation.