1. If we plug in $ x = a$ and $ y = 0$ we get $ 2f(a) = 2(f(a)+f(0))$ therefore $ f(0) = 0$. 2. If we plug in $ x = 0$ and $ y = a$ we get $ f(a)+f(-a) = 2(f(a)+f(0))$ so $ f(a)+f(-a) = 2f(a)$ hence $ f(a) = f(-a)$, therefore $ f(x) = f(-x)$ for any $ x\in\mathbb{Q}$. 3. Take any $ t\in\mathbb{N}$. Then, plugging into the original equation $ x = y =\frac{1}{t}$ we obtain $ f(\frac{2}{t})+f(0) = 2(f(\frac{1}{t})+f(\frac{1}{t}))$ hence $ f(\frac{2}{t}) = 4f(\frac{1}{t})$. Let us now prove by mathematical induction that for any $ n\in\mathbb{N}$ have $ f(\frac{n}{t}) = n^{2}f(\frac{1}{t})$. Base Step: For $ n = 1,2$ the result is true -- we have proved $ f(\frac{2}{t}) = 4f(\frac{1}{t})$ in the first paragraph of 3, and also $ f(\frac{1}{t}) = 1^{2}f(\frac{1}{t})$ Induction Step: Let us assume the result is true for $ n = k,k-1$. Then, let us prove the result for $ n = k+1$. If we plug into the original equation $ x =\frac{n}{t}$ and $ y =\frac{1}{t}$. Then we get: $ f(\frac{k+1}{t})+f(\frac{k-1}{t}) = 2(f(\frac{k}{t})+f(\frac{1}{t}))$ $ \Leftrightarrow f(\frac{k+1}{t}) = 2f(\frac{k}{t})+2f(\frac{1}{t})-f(\frac{k-1}{t})$ $ \Leftrightarrow f(\frac{k+1}{t}) = 2k^{2}f(\frac{1}{t})+2f(\frac{1}{t})-(k-1)^{2}f(\frac{1}{t})$ $ \Leftrightarrow f(\frac{k+1}{t}) = 2k^{2}f(\frac{1}{t})+2f(\frac{1}{t})-(k-1)^{2}f(\frac{1}{t})$ $ \Leftrightarrow f(\frac{k+1}{t}) = (k+1)^{2}f(\frac{1}{t})$ so the result is true for $ n = k+1$ and by mathematical induction it follows that for any $ n,t\in\mathbb{N}$ have $ f(\frac{n}{t}) = n^{2}f(\frac{1}{t})$. 4. Now, take any $ t\in\mathbb{N}$. From 3, we get $ f(1) = f(\frac{t}{t}) = t^{2}f(\frac{1}{t})$ hence $ f(\frac{1}{t}) =\frac{1}{t^{2}}f(1)$. Therefore, combining this with 3 we get for any $ n,t\in\mathbb{N}$ ${ f(\frac{n}{t}) = (\frac{n}{t})^{2}}f(1)$, and therefore for any $ x\in\mathbb{Q}$ with $ x > 0$ we get $ f(x) = x^{2}f(1)$. Finally because from 2, $ f(x) = f(-x)$, $ f(0) = 0$ and for any $ x\in\mathbb{Q}$ with $ x > 0$ we get $ f(x) = x^{2}f(1)$, then for any $ x\in\mathbb{Q}$ have $ f(x) = x^{2}f(1)$. Solution: $ f(x) = x^{2}*c$ where $ c$ is any rational number.

Plugging $x=y=0$ yields $f(0)=0.$ By induction $f(xy)=y^2f(x).$ Noting that $f$ is even we get $f(x)=cx^2,$ for some rational constant $c$, it clearly works.