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The equation $X^2 −13Y^2 = 7$ has a integer solution, such that there is a pair $(x,y) \in\Bbb Z^2$ such that $x^2 −13y^2 = 7$.

I'm struggling to understand how this is supposed to be proved without brute forcing? I am relatively a beginner to mathematical proofs, so I would like to know how you would solve a proof like this as well.

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    $\begingroup$ Grammatically? Not sure what you mean by that word here. Proving there is a solution to an equation like this can be done as simply as showing a solution. Proving there is no solution can be harder. Try looking modulo a number. $\endgroup$ Jul 11 at 16:35
  • $\begingroup$ You are probably expected to first look at suitable congruences and divisibility results. Like $7\mid x^2+y^2$ if and only if _____? Quadratic residues are another recurring trick. $\endgroup$ Jul 11 at 16:36
  • $\begingroup$ The equation can be rearranged as $X=\sqrt{7+13Y^2}$. I wrote a short Python script to test for all $Y \in \{-10^7,10^7\}$, whether or not $X$ is a whole number. The program returned no whole solutions for $X$. The closest it ever got to a whole number was for $Y = \pm 5097243, X = 18378371.000000082$ $\endgroup$
    – Daniel P
    Jul 11 at 16:52
  • $\begingroup$ Using modulo was the approach I was looking for. When I said grammatically I meant without needing to brute force a solution and prove it in that manner, apologies for the confusion. $\endgroup$
    – Spice
    Jul 11 at 22:34

3 Answers 3

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Easier than $\bmod 7$ is $\bmod 13$, whereby $x^2-13y^2=7 \Rightarrow x^2\equiv 7 \bmod 13$. The squares $\bmod 13$ are $\{0,1,3,4,9,10,12\}$. Hence there can be no integer solution to the original equation.

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  • $\begingroup$ Good point, I tried the wrong choice first! $\endgroup$
    – PC1
    Jul 11 at 17:03
  • $\begingroup$ Thank you, the example using modulus is very helpful since I was somewhat lost, @PC1's comment was helpful as well for giving a baseline which helps me get a feel for how to do this problem. $\endgroup$
    – Spice
    Jul 11 at 22:29
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We can prove that $x^2\equiv\{0,1,2,4\}\mod7$, for all $x\in\mathbb Z$. We can also prove that $13y^2\equiv\{0,3,5,6\}\mod7$ for all $y\in\mathbb Z$. So the only solutions happen when $x^2\equiv13y^2\equiv0\mod7$. This happens if and only if $x\equiv y\equiv0\mod7$.

So the left side of $x^2-13y^2$ must be a multiple of $7^2$, which is not possible as it must be equal to $7$.

There is no solution.

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This is a generalized Pell's Equation and their theory is very heavy.

For this equation, let's work modulo 7.

Then the equation is $x^2+y^2=0$ modulo 7.

$x^2$ can be 0,1,2,4 modulo seven but then $y^2$ must be 0,6,5, or 3 modulo 7 respectively. So $x$ and $y$ must be both divisible by 7.

But then from the integral equation we get $u^2-13v^2=\frac{1}{7}$ for some integers $u,v$. Absurd. So there is no solution.

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