$n=0:f(m)=f(0)+m(g(m)+g(0))$ $mg(m)-ng(n)+(m-n)g(0)=(m-n) (g(m)+g(n))$ $(m-n)g(0)=mg(n)-ng(m)$ $n=1: (m-1)g(0)=mg(1)-g(m)$ $g(m)=mg(1)-(m-1)g(0) \to g(m)=am+b$ $f(m)=f(0)+m( am+2b)=am^2+2bm+c$ where $a,b,c$ are consts Answer $(f(n),g(n))=(an^2+2bn+c,an+b)$

$$(m+1)g(m+1)=g(1)+mg(m+2)$$

Let $P(m,n)$ denote the given assertion. Then $P(m,1)+P(2,m)$ gives $f(2)-f(1)=g(m)+m(g(1)-g(2)) \implies g(m)=am+b$ where $a,b$ are constants. Hence $f(m)-f(n)=(m-n)(a(m+n)+2b) \implies f(m)-am^2-2bm=f(n)-an^2-2bn=c$ where $c$ is a constant. Hence $f(x)=ax^2+2bx+c, g(x)=ax+b \quad \forall x \in N$.