jasperE3 wrote: Find all functions $f:\mathbb R\to\mathbb R$ such that (i) $f(0)=1$; (ii) $f(x+f(y))=f(x+y)+1$ for all real $x,y$; (iii) there is a rational non-integer $x_0$ such that $f(x_0)$ is an integer. Let $x_0=\frac pq\notin\mathbb Z$ for some integers $p,q$ and let $f(\frac pq)=r\in\mathbb Z$ Let $g(x)=f(x)-1$ so that problem is : i) becomes $g(0)=0$ ii) becomes assertion $P(x,y)$ : $g(x+g(y)+1)=g(x+y)+1$ iii) becomes $g(\frac pq)=r-1$ $P(x,0)$ $\implies$ $g(x+1)=g(x)+1$ so that : $g(n)=n$ $\forall n\in\mathbb Z$ $P(x-y,y)$ becomes $g(x+g(y)-y)=g(x)$ and so : $g(x+n(g(y)-y))=g(x)$ $\forall x,y\in\mathbb R$ and $\forall n\in\mathbb Z$ And so $g(n(g(x)-x))=0$ $\forall x\in\mathbb R$ and $\forall n\in\mathbb Z$ Setting there $x=\frac pq$ and $n=q$, this becomes $g(qr-q-p)=0$ and so $qr-p-q=0$ (since $g(n)=n$ $\forall n\in\mathbb Z$) Which is $\frac pq=r-1\in\mathbb Z$, contradiction And so $\boxed{\text{No such function}}$