Let $x, y >1$ be integers satisfying $2x^2-1=y^{15}$. How can I prove that $5 \mid x$?

  • $\begingroup$ Hint : y^10 - y^5 +1 can`t be square for y>1 . $\endgroup$ – Mahan Apr 2 '11 at 12:59

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.


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.


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 $5|x$.

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    $\begingroup$ Isn't there an omission, when multiplying the residues {0,1,4} mod 5 by 2 and subtracting 1? I get three possibilities, not just the two you mention. $\endgroup$ – hardmath Apr 2 '11 at 15:25
  • $\begingroup$ @Ben: Hello and thanks for your time. But as hardmath stated, you have missed some cases. (and anyways I do not think that this ne can be solved using this method, since there are so many cases that you should check, and some of them are very hard to disprove, or they are not correct). $\endgroup$ – Amir Hossein Apr 2 '11 at 16:06
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    $\begingroup$ A purely modular arithmetic argument isn't gonna work, since it is true for x=1. $\endgroup$ – Thomas Andrews Apr 2 '11 at 18:19
  • $\begingroup$ @Byron: thanks! I have posted the general problem here: math.stackexchange.com/q/30935/6715. $\endgroup$ – Amir Hossein Apr 4 '11 at 18:07

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