Is this solution correct for equation $3x^2-4y^2=13$?

Question

Prove equation $3x^2-4y^2=13$ has no integer solution.

Solution: Suppose (y, 3)=1, We have:

$y≡( 1, 2) \mod (3)$

⇒ $4y^2≡( 1, 2) \ mod(3)$

$3x^2≡0 \mod (3)$

The common remainder between $3x^2$ and $4y^2$ is 0 , so we may write:

⇒ $3x^2-4y^2≡ 0 \ mod(3)$

But, $13≡1 \mod (3)$

Hence this equation has no integer solution. Similar result comes out when we consider remainder of nomials on 4.

Is this solution correct?

Answer 1

If $3 \mid y$, then $3 \mid -4y^2$ and since $3 \mid 3x^2$, you would have $3 \mid 13$, which is not true. Thus, as you stated, $3 \not\mid y$, which means $y^2 \equiv 1 \pmod 3$.

As such, $3x^2 - 4y^2 \equiv -y^2 \equiv 2 \pmod 3$. Note, you made a mistake when you stated "common remainder between $3x^2$ and $4y^2$ is $0$" (I'm not quite sure how you determined this) giving $3x^2 - 4y^2 \equiv 0 \pmod 3$. Also, $4y^2 \equiv 1 \pmod 3$ only, i.e., $4y^2 \not\equiv 2 \pmod 3$, which you indicated is a possible solution. Regardless, you have $13 \equiv 1 \pmod 3$. Since this doesn't match the left hand side, it means there are no integer solutions to $3x^2 - 4y^2 = 13$. This is basically what J. W. Tanner's question comment states.

As you state, you can also prove there are no solutions by using modulo $4$ instead. In this case, $4 \not\mid x$ (since, otherwise, you would need $4 \mid 13$ which is not true), so $x^2 \equiv 1 \pmod 4 \implies 3x^2 \equiv 3 \pmod 4$, so $3x^2 - 4y^2 \equiv 3 \pmod 4$, but $13 \equiv 1 \pmod 4$. As these congruence values don't match, it means there's no integer solution.

Answer 2

Here is another method:

$3x^2-4y^2=13$

Clearly x must be odd; let $x=2k+1$, then:

$3x^2-4y^2=12k^2+12k+3-4y^2=4(3k^2+3k-y^2)=10$

⇒$2(3k^2+3k-y^2)=5$

LHS is even but RHS is odd, so the equation can not have integer solution.

Answer 3

I think you are trying to say:

If you assume $\gcd(y,3) =1$ (Why are you assuming this? What if $\gcd(y,3) = 3$?) or in other words if $3\not\mid y$ then $y\equiv 1 \pmod 3$ or $y\equiv 2 \pmod 3$.

ANd therefore $y^2\equiv 1 \pmod 3$ or $y^2\equiv 4\equiv 1\pmod 3$. and so $4y^2 \equiv 1\pmod 3$. I'm not sure why you didn't eliminate $4y^2 \equiv 2\pmod 3$.

And $3x^2 \equiv 0\pmod 3$.

Then you said "The common remainder between $3x^2$ and $4y^2$ is $0$" and I can not figure out what you mean and I can't think of any interpretation that would be compatible with you discovery that $3x^2 \not \equiv 4y^2$.

You claim $3x^2 - 4y^2 \equiv 0\pmod 3$ which is completely wrong as $4y^2 \equiv 1\pmod 3$ we have $3x^2 - 4y^2 \equiv -1 \equiv 2 \pmod 3$.

Which is enough to show $13\equiv 1\not \equiv 2\pmod 3$. So there is not integer solution.

IF we assume $\gcd(3,y)=1$. But you didn't explain why we are assuming $\gcd(3,y) = 1$.

Which you never considered.

Better to simply note:

$3x^2 - 4y^2 \equiv \begin{cases}3*0^2\equiv 0& x\equiv 0 \\3*1^2\equiv3\equiv 0 &x\equiv 1\pmod 3\\3*2^2\equiv 3*4\equiv 0&x\equiv 2\pmod 3\end{cases}-\begin{cases}4*0^2\equiv 0& y\equiv 0\pmod 3\\4*1^2\equiv 1&y\equiv 1\pmod 3\\4*2^2\equiv 16\equiv 1&y\equiv 2\pmod 3\end{cases}\pmod 3\equiv$

$0 - \begin{cases}0& \text{if }y\equiv 0\pmod 3\\1&\text{otherwise}\end{cases}\equiv$

$\begin{cases}0\\-1\end{cases}\equiv \begin{cases}0\\2\end{cases}\pmod 3$

$\not \equiv 1 \equiv 13 \pmod 3$.

And it's enough to prove it for just modulo $3$.

FWIW if you tried modulo $4$ you'd get.

$3x^2 - 4y^2 \equiv 3x^2\equiv 3*\begin{cases}0^2\\1^2\\2^2\\3^2\end{cases}\equiv$

$\begin{cases}0\\3\\0\\3\end{cases}\not \equiv 1\equiv 13\pmod 4$.

But there was utterly no reason to do those as we were done.