$x,y$ are integers satisfying $2x^2-1=y^{15}$, show that $5 \mid x$ Let $x, y >1$ be integers satisfying $2x^2-1=y^{15}$. How can I prove that $5 \mid x$?
 A: You can rewrite this as $1-2x^2 = (-y)^{15}$ and then use unique factorization in the ring $\mathbb{Z}[\sqrt{2}]$ to show that $1+x\sqrt{2}$ must be a 15th power in the ring.  Then write
$$1+x\sqrt{2} = u (a+b\sqrt{2})^{15}$$
Where $u$ is a unit of the ring.
A: Here is a proof that works even for the fifth power.
Suppose $2x^2-1=y^5$. Then we can write $2x^2=(y+1)(1-y+y^2-y^3+y^4)$.
Writing 
$$5=(1-y+y^2-y^3+y^4)+(10(y+1)-10(y+1)^2+5(y+1)^3-(y+1)^4)$$
shows that $g(y):=\gcd(y+1,1-y+y^2-y^3+y^4)$ is always equal to 1 or 5.
If $g(y)=5$, then 5 divides $2x^2$, hence 5 divides $x$, and we are done.
The other case leads to a contradiction.
If $g(y)=1$, then  $y+1$ and $1-y+y^2-y^3+y^4$ are relatively
prime, and their product is twice a square. Thus, one of them is a 
square and the other is twice a square. Since $1-y+y^2-y^3+y^4$ is odd, it must be a square. 
But for $y>1$, you can easily check that 
$$(2y^2-y)^2 < 4(1-y+y^2-y^3+y^4) <(2y^2-y+1)^2,$$
so $1-y+y^2-y^3+y^4$ is not a square. 
(I learned this from Ed Burger's book "Exploring The Number Jungle").
This is the required contradiction. 

Added: I'm pretty sure that there are no integer solutions to $2x^2-1=y^5$
with $y>1$, but I can't prove it yet. 
A: Basically any number in the universe can be written in the form $5x$, $5x+1$, .... $5x+4$. If I square any of these numbers, I get either a number of the form $5x$, $5x+1$, $5x+4$, where $x$ is an integer. So 2 multiplied by any of these numbers, then taking away 1 is of the form $5x+1$, $5x+2$, where $x$ again is another integer. So as the right hand side is $y^{14}(y)$, we have that it is an integer of the form $5x$ or $5x+1$ times some integer of the form $5a$, $5b+1$....$5b+4$, and so $y^{15}$ is of the form....?
Complete the argument as an exercise, and you'll see why $5|x$.
