Problem
ALG-B3-M02-P009 Difference of quadratic type
#9
★★★★☆ Level 4 of 5
Let \(f:\mathbb Q\to\mathbb Q\), \(f(x+y)-f(x-y)=4xy\), \(f(0)=0\), \(f(1)=1\). Find \(f\).
Hint. Check \(f(q)=q^2\), then subtract \(q^2\).
Put \(g(x)=f(x)-x^2\). Then \(g(x+y)=g(x-y)\) for all \(x,y\). Choosing \(u=x+y\), \(v=x-y\), we can represent any rational \(u,v\) by \(x=(u+v)/2\), \(y=(u-v)/2\). Hence \(g(u)=g(v)\) for all \(u,v\), so \(g\) is constant. Since \(g(0)=0\), \(g\equiv0\), and \(f(q)=q^2\).
Goal: choose the right first substitution and then verify the found function.