Problem
ALG-B1-M08-P023 Quadratic Cauchy substitution
#23
★★★★★ Level 5 of 5
Let \(f:\mathbb Q\to\mathbb Q\) satisfy \(f(x+y)=f(x)+f(y)+2xy\) for all \(x,y\in\mathbb Q\), and \(f(1)=1\). Find \(f\).
Subtract \(x^2\): set \(g(x)=f(x)-x^2\).
Let \(g(x)=f(x)-x^2\). Then \(g(x+y)=f(x+y)-(x+y)^2=f(x)-x^2+f(y)-y^2=g(x)+g(y)\). Thus \(g\) is additive on \(\mathbb Q\).
Since \(g(1)=f(1)-1=0\), \(g(q)=0\) for all rational \(q\). Therefore \(f(x)=x^2\). The check is immediate: \((x+y)^2=x^2+y^2+2xy\).
This is the main template of “subtract the known part.”