Hint: Let $g(x)=f(x)+1$, we get $g(xy)+g(x+y)=g(x)g(y)+1$. We can demonstrate by induction and NZQR method, that $g(x)=x+1$. PS: I sendt you a PM with the complete solution.

Mr.Math.Boy wrote: Hint: Let $g(x)=f(x)+1$, we get $g(xy)+g(x+y)=g(x)g(y)+1$. We can demonstrate by induction and NZQR method, that $g(x)=x+1$. You forgot $g(x)\equiv 1$

Quidditch wrote: Find all functions $f:\mathbb{Q}\to\mathbb{Q}$ such that $$f(xy)+f(x+y)=f(x)f(y)+f(x)+f(y)$$for all $x,y\in\mathbb{Q}$. Let $P(x,y)$ be the assertion $f(xy)+f(x+y)=f(x)f(y)+f(x)+f(y)$ Let $c=f(1)$ If $c=0$, $P(x-1,1)$ $\implies$ $\boxed{\text{S1 : }f(x)=0\quad\forall x\in\mathbb Q}$, which indeed fits If $c\ne 0$ : $P(0,1)$ $\implies$ $f(0)=0$ $P(1,1)$ $\implies$ $f(2)=c^2+c$ $P(2,1)$ $\implies$ $f(3)=c^3+c^2+c$ $P(3,1)$ $\implies$ $f(4)=c^4+c^3+c^2+c$ $P(2,2)$ $\implies$ $f(4)=\frac{c^4+2c^3+3c^2+2c}2$ Subtracting these two last lines, we get $c^4=c^2$ and so $c\in\{-1,1\}$ If $c=-1$ $P(x,1)$ $\implies$ $f(x+1)=-f(x)-1$ and so $f(x+2)=-f(x+1)-1=f(x)$ So $f(2)=f(0)=0$ and $P(x,2)$ implies $f(2x)=0$ $\forall x$, impossible here So no solution there and $c=1$ $P(x,1)$ $\implies$ $f(x+1)=f(x)+1$ and so $f(x+n)=f(x)+n$ $\forall x\in\mathbb Q$ and $\forall n\in\mathbb Z$ $P(x,n)$ $\implies$ $f(nx)=nf(x)$ And so $f(\frac pq)=\frac pqf(1)$ And so $\boxed{\text{S2 : }f(x)=x\quad\forall x\in\mathbb Q}$, which indeed fits

pco wrote: Mr.Math.Boy wrote: Hint: Let $g(x)=f(x)+1$, we get $g(xy)+g(x+y)=g(x)g(y)+1$. We can demonstrate by induction and NZQR method, that $g(x)=x+1$. You forgot $g(x)\equiv 1$ Thank you, you are right!