Note that $Q(x)$ cannot have two distinct real roots

DavyDuf wrote: Find all Polynomial $P(x)$ and $Q(x)$ with Integer Coefficients satisfied the equation: \[Q(a+b) = \frac{P(a) - P(b)}{a - b}\]$\forall a, b \in \mathbb{Z}^+$ and $a>b$ This is $(a-b)Q(a+b)=P(a)-P(b)$ $\forall a>b\in\mathbb Z^+$ and so, since polynomials, $\forall a,b\in\mathbb R$ So $(a-b)Q(a+b)+(b-c)Q(b+c)+(c-a)Q(c+a)=0$ Choosing then $a+b=x,b+c=0,c+a=1$ and so $(a,b,c)=(\frac{x+1}2,\frac{x-1}2,\frac{1-x}2)$, this becomes $Q(x)+(x-1)Q(0)-xQ(1)=0$ and so $Q(x)=ux+v$ for some $u,v\in\mathbb Z$ and so $P(x)=ux^2+vx+w$

On phone All deg P<=2 work and the corresponding Q is obvious. Now note that this equality should hold as a polynomial identity too. On the other hand if we divide a^n-b^n from P(a)-P(b) by a-b we get some homogeneous degree n-1 polynomial. The only way we get degree n-1 terms from Q is (a+b)^(n-1). Clearly they’re only equal when n<=2, so the conclusion follows