Solve for all $x,y\in N$ and $x^3-3^y = 2015$

Question

Question:

Solve for all $x,y\in N$ and $x^3-3^y = 2015$

So far I have tried $\mod(7)$ and $\mod(8)$ but I am not sure if I am in the right path or not I will be happy if someone can toss me a hint.

Answer 1

Use Metin's idea of "reject solution of too large" and someone's idea of mod $13$ -> $3\mid m$, I reach a solution.

Consider mod $13$, then

$3^y\equiv1,3,9 \pmod {13}$

$x^3\equiv1,5,8,12 \pmod {13}$

Thus, $3^y\equiv1 \pmod {13}, x^3\equiv1 \pmod {13}, 3\mid y$

Let $y=3z, k=3^z$, so $x>k>0$

Then $x^3-3^y = x^3-(3^z)^3 = x^3-k^3 = (x-k)(x^2+xk+k^2) = 2015 $

$x^2+xk+k^2$ is factor of $2015$ and thus bounded above, we can have $x<45$. Then the rest is check and reject by computer. (By hand is tough)

Furthermore, consider $x^3-27^z=2015$ with mod $7$, we have $7\mid x$. (Consider all outcomes as above). With $x\equiv2 \pmod{6}$ by Alexander. We can reject all $x<45$ except $14$.

Answer 2

EDIT: The proof has a fatal flaw as pointed out in the comments.

We will show that the solutions cannot be infinitely large and reduce the infinite search space to a finite one. This basically solves the problem because we can check the remaining finite number of candidates either by hand or by a computer.

Assume $(x, y)$ is a solution. Of course we have $3^y < x^3$. Also, for $x \geq 27$, we have $x^3 - (x-1)^3 > 2015$; therefore, we need to have $(x-1)^3 < 3^y < x^3$. This is equivalent to $\frac{3}{\log 3} \log (x-1) < y < \frac{3}{\log 3} \log (x)$. However, notice that $\frac{3}{\log 3} \log (x) - \frac{3}{\log 3} \log (x-1) < 1$ for $x \geq 4$, so there can be no integer between them.

This shows that there is no solution when $x \geq 27$.

Searching through the numbers with $x < 27$ and $y < 9$, we find that the only solution is $x = 14$ and $y = 6$.