Exercise 1.5.6.

Question

Prove that the Diophantine equation $x^2 - 5y^2 = 3$ has no solution.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

\begin{proof}
If $(x,y) \in \Z^2$ satisfies $x^2 - 5y^2 = 3$, then $x^2 \equiv 3 \pmod 5$.

But for all $x \in \Z$, $x \equiv 0, \pm 1, \pm 2$, so $x^2 \equiv 0, 1 ,4 \pmod 5$, and $3$ is not congruent to $0, 1, 4$ modulo $5$.

This contradiction shows that $x^2 - 5y^2 = 3$ has no integer solution.
\end{proof}

Answers