How to prove the Diophantine equation : $x^2 - 71y^2 = -1$ has no solution with x,y in the set of integers

Question

How to prove the Diophantine equation : $x^2 - 71y^2 = -1$ has no solution with $x,y$ in the set of integers?

I am stuck on this question for a long time and i cant really understand how to prove it. Can someone help please?

Answer 1

A solution to a Pell-type equation

$$x^2 - Dy^2 = a\tag{1}$$

gives in particular a solution to the congruence

$$x^2\equiv a \pmod{D}.\tag{2}$$

So a necessary condition for the solubility of $(1)$ is that $a$ be a quadratic residue modulo $D$, hence modulo every prime dividing $D$.

It is easy to see that $x^2 \equiv -1 \pmod{p}$ has no solution if $p$ is a prime $\equiv 3 \pmod{4}$, since for $p > 2$, the congruence $x^2\equiv -1\pmod{p}$ implies that the multiplicative order of $x$ modulo $p$ is $4$, hence $4 \mid (p-1)$, or $p \equiv 1 \pmod{4}$. Since $71 \equiv 3 \pmod{4}$, there is no solution to $x^2 \equiv -1 \pmod{71}$, and a fortiori no solution to $x^2 - 71 y^2 = -1$.

Thus a necessary condition for the existence of solutions to

$$x^2 - Dy^2 = -1$$

is that $D$ has no prime divisors $\equiv 3\pmod{4}$, and that $D$ is not divisible by $4$, and not a perfect square (if $D > 1$). But these conditions are not sufficient, for example $x^2-34y^2 = -1$ and $x^2 - 221y^2 = -1$ have no solutions although $34 = 2\cdot 17$ and $221 = 13\cdot 17$ satisfy these conditions.

It is less easy to see, but true, that for every prime $p \equiv 1 \pmod{4}$ the equation $x^2 - p y^2 = -1$ has solutions.