We first prove that there exist infinitely many pairs of rational numbers $(x,y)$ such that $$x^2+y^2 = 2.$$Take any Pythagorean triple $(a,b,c)$ with $b > a$ (note that there are infinitely many such Pythagorean triples). Then we can take $x = \dfrac{b-a}{c},y = \dfrac{b+a}{c}$, which gives $x^2+y^2 = 2$. Now take any positive integer $n$. If $n$ is even, then take $\dfrac{n}{2}$ pairs of rationals with sum of squares equal to $2$. If $n$ is odd, then take $\dfrac{n-1}{2}$ such pairs, and one number $1$. We now prove that all the rationals are distinct. If $\dfrac{b+a}{c} = \dfrac{b-a}{c}$, then $a = 0$, an impossibility. Thus we consider two different primitive Pythagorean triples, $(a,b,c)$ and $(a',b',c')$ (with $c < c'$) that generate at least two of the same rational numbers. Since the sum of squares of those rationals are equal to $2$, the other pair of rationals must be equal. Thus we have either \begin{align*}\dfrac{b-a}{c} &= \dfrac{b'-a'}{c'} \quad \text{and} \quad \dfrac{b+a}{c} = \dfrac{b'+a'}{c'} \implies \dfrac{b-a}{b'-a'} = \dfrac{b+a}{b'+a'} = \dfrac{c}{c'} = \lambda \in \mathbb{Q}, \text{ }\text{or}\\\dfrac{b-a}{c} &= \dfrac{b'+a'}{c} \quad \text{and} \quad \dfrac{b+a}{c} = \dfrac{b'-a'}{c'} \implies \dfrac{b-a}{b'+a'} = \dfrac{b+a}{b'-a'} =\dfrac{c}{c'} = \lambda \in \mathbb{Q}.\end{align*}In both cases we have $a^2+b^2 = c^2 = \lambda^2(c')^2 = \lambda^2((a')^2+(b')^2)$ and $b^2-a^2 = \lambda^2((b')^2-(a')^2)$. Thus $c^2 = \lambda^2(c')^2,a^2 = \lambda^2(a')^2,b^2 = \lambda^2(b')^2$. But then since $\lambda = \dfrac{p}{q}$ where $\gcd(p,q) = 1$, $q \mid a',b',c'$, which contradicts the fact that our triples are primitive. Thus all our rationals are distinct.

We first prove that there exist infinitely many pairs of rational numbers $(x,y)$ such that $$x^2+y^2 = 2.$$Take any Pythagorean triple $(a,b,c)$ with $b > a$ (note that there are infinitely many such Pythagorean triples). Then we can take $x = \dfrac{b-a}{c},y = \dfrac{b+a}{c}$, which gives $x^2+y^2 = 2$. Now take any positive integer $n$. If $n$ is even, then take $\dfrac{n}{2}$ pairs of rationals with sum of squares equal to $2$. If $n$ is odd, then take $\dfrac{n-1}{2}$ such pairs, and one number $1$. We now prove that all the rationals are distinct. If $\dfrac{b+a}{c} = \dfrac{b-a}{c}$, then $a = 0$, an impossibility. Thus we consider two different primitive Pythagorean triples, $(a,b,c)$ and $(a',b',c')$ (with $c < c'$) that generate at least two of the same rational numbers. Since the sum of squares of those rationals are equal to $2$, the other pair of rationals must be equal. Thus we have either \begin{align*}\dfrac{b-a}{c} &= \dfrac{b'-a'}{c'} \quad \text{and} \quad \dfrac{b+a}{c} = \dfrac{b'+a'}{c'} \implies \dfrac{b-a}{b'-a'} = \dfrac{b+a}{b'+a'} = \dfrac{c}{c'} = \lambda \in \mathbb{Q}, \text{ }\text{or}\\\dfrac{b-a}{c} &= \dfrac{b'+a'}{c} \quad \text{and} \quad \dfrac{b+a}{c} = \dfrac{b'-a'}{c'} \implies \dfrac{b-a}{b'+a'} = \dfrac{b+a}{b'-a'} =\dfrac{c}{c'} = \lambda \in \mathbb{Q}.\end{align*}In both cases we have $a^2+b^2 = c^2 = \lambda^2(c')^2 = \lambda^2((a')^2+(b')^2)$ and $b^2-a^2 = \lambda^2((b')^2-(a')^2)$. Thus $c^2 = \lambda^2(c')^2,a^2 = \lambda^2(a')^2,b^2 = \lambda^2(b')^2$. But then since $\lambda = \dfrac{p}{q}$ where $\gcd(p,q) = 1$, $q \mid a',b',c'$, which contradicts the fact that our triples are primitive. Thus all our rationals are distinct.

We first prove that there exist infinitely many pairs of rational numbers $(x,y)$ such that$$x^2+y^2 = 2.$$