Diophantine Equatiοn $x^3=2^y+15$ I would like some help with the diophantine equation
$x^3=2^y+15$
I have tried working with last digits and modular arithmetic but that hasn't got me anywhere.
 A: Hint: An unfairly neglected modulus is $7$. 
A: To begin, notice some things:


*

*$x^3 -15$ must be divisible by 2

*$x$ must be odd as well


Also, try and use the unit digits:
$2^y$ can end in:
$$2,4,6, 8$$
Case 1: If it ends in 2
Then,
$x^3$ ends in 7, which occurs if $x$ has a units digit of 3.
Case 2: If it ends in 4
Then,
$x^3$ ends in 9, which occurs if $x$ has a units digit of 9.
Case 3: If it ends in 8
$x^3$ ends in 3, which occurs if $x$ has a units digit of 7.
Case 4: If it ends in 6
$x^3$ ends in 1, which occurs if $x$ has a units digit of 1.
Therefore, we can conclude that $x$ follows the units digits of $3^n$.
NOTE
I haven't found a full solution yet, I'm just posting some hints to start you off.
A: The tempting thing is to try a suitable power of $2$ as modulus. Then for $y$ large enough you are left solving $x^3 \equiv 15 \bmod 2^m$
However, since $3$ is coprime to $\phi(2^m) = 2^{m-1}$ you will always be able to solve this congruence. Hence working mod a power of $2$ does not tell you there are no solutions to this Diophantine.
This however gives you a clue to what modulus $n$ you might like to use. We need one such that hcf$(3,\phi(n)) \neq 1$. The first possibility is $n=7$ and this works...
(Of course for the purpose of the exercise you don't have to do everything I did above, you just need to stumble upon modulus $7$...however I thought I would explain the thought processes that might get someone more familiar with this stuff there).
