19 is not the sum of two perfect rational squares Can someone show me a proof as to why 19 is not the sum of the squares of two rational number? I have seen similar proofs for other numbers but I need help with 19 specifically since its residue system is so large.
 A: Note that the square of an integer can only be $0$ or $1$ mod $4$. So the sum of two perfect squares can only be $0$, $1$ or $2$ mod $4$.
An odd integer is representable as the sum of two integers if and only if it is $1$ mod $4$.
Let $n\equiv 0$ (mod $4$). If $n=x^2+y^2$ for some $x,y\in\mathbb{N}$, then $x$ and $y$ are both even. So $\frac{n}{4}=(\frac{x}{2})^2+(\frac{y}{2})^2$.
Now suppose that $$19=\left(\frac{p}{q}\right)^2+\left(\frac{r}{s}\right)^2$$ for some $p,q,r,s\in\mathbb{N}$ with $q,s\ne0$. Then we have
$$19(qs)^2=(ps)^2+(rq)^2$$
If $qs$ is odd and $(qs)^2\equiv 1$ (mod $4$) and hence $19(qs)^2\equiv 3$ (mod $4$) , which is impossible. So $qs$ is even and hence $(qs)^2$ is divisible by $4$. So $19(\frac{qs}{2})^2$ is the sum of two perfect squares. We can repeatedly divide $qs$ by $2$ until the quotient is odd, but then the quotient is 3 (mod $4$) and cannot be the sum of two perfect square. This leads to a contradiction.
A: Because $x^2\equiv(0,1,4,9,-3,6,-2,-8,7,5)\mod19$ and there does not exist $\{a,b\}\subset\{1,4,9,-3,6,-2,-8,7,5\}$, for which $a+b$ divided by $19$ leaves no remainder.
The rest is an infinite descent.
A: A square must be $\equiv 0$ or $1$ mod $4$, hence a sum of squares must be $0,1$ or $2$ mod $4$. But $19 \equiv 3$ mod $4$.
