Modular Arithmetic. Integer Solutions

Question

How can you show that $|a^2 -10b^2|=2$ has no integer solutions for a and b using modular arithmetic? Thank you.

Answer 1

Modulo $5$, we get $$a^2\equiv 2\mod 5$$ or $$a^2\equiv 3\mod 5$$ which both is not solvable over the integers.

Answer 2

This is a case where the principal form also represents $-1,$ as $3^2 - 10 = -1.$ So, if $a^2 - 10 b^2 = n,$ we find $$ (3a+10b)^2 - 10 (a+3b)^2 = -n $$

Meanwhile, $x^2 - 10 y^2$ represents primes $p \equiv 1, 9, 31, 39 \pmod {40}$

The other class of this discriminant $2x^2 - 5 y^2,$ represents $2$ and $5,$ then primes $p \equiv 3, 13, 27, 37 \pmod {40}$

If $2a^2 - 5 b^2 = k,$ we find $$ 2(3a+5b)^2 - 5 (2a+3b)^2 = -k $$

Answer 3

Best to work modulo $10$. You know, from elementary school, that the only squares modulo $10$ are $0,1,4,5,6$, and $9$. The impossibility of $\pm2=m^2-10n^2$ follows.