Since $ f(x + 1) = f(x) + 1$, by induction we have $ f(x + n) = f(x) + n$ for all rational number x For all positive integers $ a$ and $ b$ we have: $ f((\frac {a}{b} + b)^2) = f(\frac {a^2}{b^2} + 2a + b^2) = f(\frac {a^2}{b^2}) + 2a + b^2$ $ \leftrightarrow f^2(\frac {a}{b} + b) = f(\frac {a^2}{b^2}) + 2a + b^2$ $ \leftrightarrow [f(\frac {a}{b}) + b]^2 = f(\frac {a^2}{b^2}) + 2a + b^2$ $ \leftrightarrow f(\frac {a}{b}) = \frac {a}{b}$ which implies $ f(x) = x \forall x \in Q^ +$ P/S: By the similar way, we can replace 2 by 3./.

By induction, $$f(x+k)=f(x)+k.$$Let $f(pq^{-1})=n,$ then $$f(pq^{-1}+dq)=n+dq \implies f((pq^{-1}+dq)^2)=n^2+2dnq ~~\forall d\in \mathbb{N}.$$Since $$(pq^{-1}+dq)^2=\frac{p^2}{q^2}+2pd+d^2q^2 \implies f((pq^{-1}+dq)^2)=f(\frac{p^2}{q^2})+2pd+d^2q^2=q^2+2pq+d^2q^2\implies 2dnq =2pd \implies n=pq^{-1}.$$Thus $f(x)=x ~~\forall x\in \mathbb{Q^+},$ is the only solution that indeed fits.