Problem
ALG-B3-M07-P015 Symmetric Difference
#15
★★★★★ Level 5 of 5
Find all \(P\in\mathbb R[x]\) such that \(P(x+1)-P(x-1)=4x\).
Compare with \(x^2\).
For \(x^2\), \((x+1)^2-(x-1)^2=4x\). Therefore \(Q=P-x^2\) satisfies \(Q(x+1)=Q(x-1)\), i.e. \(Q(x+2)=Q(x)\). A periodic polynomial is constant. The answer is \(P(x)=x^2+c\).
Symmetric finite difference.