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I want to use modular congruences to show that $5=3x^2-4y$ has no solutions on $(x,y)\in\mathbb Z^2$. Here's my attempt:

Take $3x^2\equiv_5 4y$. Then since $2$ is an inverse modulo $5$, $x^2\equiv_58y\equiv_53y$. Hence, $$x^4\equiv_5 4y^2$$ By Fermat's little theorem, since $5$ is prime we require $x^4\equiv_5 1$. However, $4y^2\equiv_5 1$ iff $y^4\equiv_5 1$. So, $x^4\equiv_5y^4$.

Here I'm stuck on how to go further.

Bill Dubuque
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Lex_i
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    I think there's a much easier way to go about this: try considering the residues mod $4$ instead, it follows very directly that there are no solutions – Stephen Donovan May 26 '22 at 07:53
  • The common way is to reduce if $\bmod 4$ which yields $, 1\equiv -x^2$ so $,x^2\equiv -1\equiv 3,,$ contra the linked dupe. – Bill Dubuque May 26 '22 at 08:01
  • So what I've done using the first suggestion is write $3x^2\equiv_4 5$, which implies $x^2\equiv_4 3$. Used this formula in addition to $5=3x^2-4y$ to create a matrix and rref the matrix. This gave only one solution $(1, -1/2)$ which is obviously not in $\mathbb Z^2$. That should be good, right? – Lex_i May 26 '22 at 08:03

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