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Prove that there are no integers $x$ and $y$ such that $3x^2=13+4y^2$.

From the equation, I know that $3x^2$ must be odd and therefore equal $2k + 1$ for some integer $k$.

But I am unsure what to do after that.

I have also worked out that $k = 2(y^2 + 3)$, but I don't know if that helps at all.

My instructor noted that I should look at whether $x^2$ is even or odd, but I am at a loss.

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    $\begingroup$ x must be odd. Because if x is even so is $x^2$ and so is $3x^2$ but $13+4y^2$ is odd $\endgroup$ – Isham Sep 18 '17 at 22:25
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$3x^2=\underbrace{13+4y^2}_\text{odd}$

We know that x is odd lets' pose $x=2k+1$

\begin{align}3(2k+1)^2=13+4y^2\\ 3(4k^2+4k+1)=13+4y^2\\ 3(4k^2+4k)=10+4y^2\\ 12(k^2+k)=10+4y^2\\ \underbrace{6(k^2+k)}_\text{even}=\underbrace{5+2y^2}_\text{odd} \end{align}

but $5+2y^2$ is odd not even which contradict the factor 6 on the left side of the equation

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You can look mod4.

Your equation is equivalent to

$1=3x^{2} \pmod 4$.

$3=x^{2}\pmod 4$.

But 3 is not a square mod 4.

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Square of an integer is always of the form $4k$ or $4k+1$.

$$\text{RHS} =4(y^2+3)+1=4n+1$$

While, on the other hand (left one! :P)

$$\text{LHS}=3(4m) ~\text{or} ~3(4m+1) \equiv 4l ~ \text{or} ~4l+3$$

LHS and RHS both leave different remainders while dividing with $4$, therefore they can never be equal.

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Ignore $4y^2$ for a minute.

The equation $3x^2 = 13$ has no solutions in integers either. If it had a solution, then $x$ would have to be odd, obviously. Then $3x^2 \equiv 3 \bmod 4$ but $13 \equiv 1 \bmod 4$.

Since $4y^2 \equiv 0 \bmod 4$, the equation $3x^2 = 4y^2 + 13$ is likewise hopelessly insoluble (is that the right word? you know what I mean) in integers.

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Welp, you know that $x$ must be odd. Let's chase our tail.

So let $x = 2k+1$ and

$12k^2 + 12k + 3 = 13 + 4y^2$

$12k^2 + 12k = 10 + 4y^2$.

$6k^2 + 6k = 5 + 2y^2$ which isn't possible.

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