Find all positive integers n such that 3x^2-y^2=2018^n has an integer solution

[littlemathquark]

Homework Statement

Find all positive integers $n$ such that $3x^2-y^2=2018^n$ has integer solutions.

Relevant Equations

None

I need an idea. Thank you.

 

[fresh_42]

If $n$ is even consider the equation modulo $3.$

The hard part is to show that odd $n$ actually do have solutions.

 

[littlemathquark]

$3x^2-y^2\equiv 2018^n\mod 3$ $$-y^2\equiv 2^n\equiv 1\mod 3$$ $$y^2\equiv 2\mod 3$$ but it is impossible so $n$ must be odd.

 

[fresh_42]

$3x^2-y^2\equiv 2018^n\mod 3$ $$y^2\equiv 2^n\equiv 1\mod 3$$

This is wrong. Use the assumption that $n$ is even.

(For odd $n$ show that $3x^2-x^2\equiv 2018^n$ has always a solution if $3x^2-x^2\equiv 2018$ has a solution. So you will be left with the case $n=1.$)

 

[littlemathquark]

I mean if n is even $y^2\equiv 2\mod3$ but it is impossible so n must be odd.

 

[fresh_42]

I mean if n is even $y^2\equiv 2\mod3$ but it is impossible so n must be odd.

Right.

Now show that if $3a^2-b^2=2018$ then there will be a solution for all $3y^2-x^2=2018^n$ if $n$ is odd.

 

[littlemathquark]

$$3x^2=y^2+2018\implies x^2\geq \frac{2018}{3}\implies x\geq 26.$$ For $x=27$ , $y=13$ satisfy equation.

 

[fresh_42]

$$3x^2=y^2+2018\implies x^2\geq \frac{2018}{3}\implies x\geq 26.$$ For $x=27$ $y=13$

That's not what I said. This will be the third and last step but first, make the second. What does it mean that $n$ is odd?

 

[littlemathquark]

If n is odd then $n=2k+1$

 

[fresh_42]

If n is odd then $n=2k+1$

Yes. Go ahead. How can you corstruct a solution for $3x^2-y^2=2018^{2k+1}=2018\cdot (2018^{k})^2$ if you have a solution $3a^2-b^2=2018$?

 

[littlemathquark]

$3a^2.2018^{2k}-b^2.2018^{2k}=2018^{2k+1}$

 

[fresh_42]

$3a^2.2018^{2k}-b^2.2018^{2k}=2018^{2k+1}$

Yes. I would have written it as
$$
3x^2-y^2=2018^n=2018\cdot 2018^{2k}=
(3a^2-b^2)\cdot 2018^{2k}=3\cdot (a\cdot 2018^k)^2 - (b\cdot 2018^k)^2
$$
but yes.

So what is the answer to your initial question then?

 

[littlemathquark]

$3a^2.2018^{2k}-b^2.2018^{2k}=2018^{2k+1}$ İf $a=27$ and $b=13$ then $x=(27.2018^k)$ and $y=(13.2018^k)$ are solutions. Thank you very much. But are these all solutions? Are there other solutions?

 

[fresh_42]

$3a^2.2018^{2k}-b^2.2018^{2k}=2018^{2k+1}$ İf $a=27$ and $b=13$ then $x=(27.2018^k)$ and $y=(13.2018^k)$ are solutions. Thank you very much. But are these all solutions? Are there other solutions?

There are infinitely many solutions. With $n,m\geq 0$ these are (for $2018$)
\begin{align*}
3x_n^2-y_n^2&=2018\\[6pt]
x_n &= \pm \dfrac{1}{6} \left(-81 (2 - \sqrt{3})^n + 13 \sqrt{3} (2 - \sqrt{3})^n - 81 (2 + \sqrt{3})^n - 13 \sqrt{3} (2 + \sqrt{3})^n\right)\\
y_n &= \pm \dfrac{1}{2} \left(-13 (2 - \sqrt{3})^n + 27 \sqrt{3} (2 - \sqrt{3})^n - 13 (2 + \sqrt{3})^n - 27 \sqrt{3} (2 + \sqrt{3})^n\right)\\[12pt]
3\hat x_m^2-\hat y_m^2&=2018\\[6pt]
\hat x_m &= \pm \dfrac{1}{6} \left(81 (2 - \sqrt{3})^m + 13 \sqrt{3} (2 - \sqrt{3})^m + 81 (2 + \sqrt{3})^m - 13 \sqrt{3} (2 + \sqrt{3})^m\right)\\
\hat y_m &= \pm \dfrac{1}{2} \left(13 (2 - \sqrt{3})^m + 27 \sqrt{3} (2 - \sqrt{3})^m + 13 (2 + \sqrt{3})^m - 27 \sqrt{3} (2 + \sqrt{3})^m\right)\\[12pt]
x_0&=\hat x_0=27\\
y_0&=\hat y_0=13\\[6pt]
x_1&=67\\
y_1&=107\\[6pt]
\hat x_1&=41\\
\hat y_1&=55\\[6pt]
\ldots \; etc.
\end{align*}
But I used Wolfram Alpha for it. I haven't found them myself. But your idea to find $(27,13)$ was nice. The solutions for higher odd powers of $2018$ are as described above:
$$
2018^{2k+1}=3(x_n\cdot 2018^k)^2-(y_n\cdot 2018^k)^2\, , \,2018^{2k+1}=3(\hat x_m\cdot 2018^k)^2-(\hat y_m\cdot 2018^k)^2
$$

 

[fresh_42]

The way we followed here was a typical one:

1.) Eliminate the impossible cases ($n=2k$) by a modulo consideration.
2.) Reduce the general case ($n=2k+1$) to the simple case ($n=1$).
3.) Solve the simple case ($(x,y)=(27,13)$).

 

[fresh_42]

I added the rest of the solutions and another two examples of low values in post #14 for completion.