Suppose P has degree n and $ P(x)=\sum_{i=1}^na_nx^n$. Obviously all $ a_n$ have to be rational. We notice that $ P(ax)+b$ holds when $ P(x)$ holds, where a and b are rationals, a is non-zero. We can assume, wlog, all $ a_n$ are integers and $ a_0=0$. Let q be a prime with $ gcd(q,a_n)=1$. To solve $ P(x)=q$, we have to have $ x=\frac yz$, where y=1 or y=q and z|a_n. Note here z can be negative. We can always find a $ z$ such that $ z|a_n$ and $ P(\frac qz)=q$ holds for infinitely many $ q$. In other words, $ P(x)-zx=0$ has infinitely many roots. Hence, $ P(x)=zx$ with $ z$ rational. Inverting the $ P(ax)+b$ operation, we know P(x)=ax+b, where a,b are rational and a is nonzero.

xxp2000 wrote: Obviously all $ a_n$ have to be rational. I don't think this is obvious at all. In particuar, we aren't given the other direction ($ d$ rational implies $ P(d)$ rational).

P(x) has degree of n. You can find n+1 distinct pairs of $ (d_i, r_i)$ such that $ P(d_i)=r_i$. then you can construct P(x) by the following formula $ P(x)=\sum_{i=1}^nr_i\frac{(x-d_1)...(x-d_{i-1})(x-d_{i+1})...(x-d_{n+1})}{(d_i-d_1)...(d_i-d_{i-1})(d_i-d_{i+1})...(d_i-d_{n+1})}$. Please let me know if it is still obvious.