Find all this diophantine equation $$24x^4+1=y^2\tag{1} $$ postive integers solution

it is clear $(x,y)=(1,5)$

I know $y^{2}=Dx^{4}+1$, where $D>0$ and is not a perfect square, has at most two solutions in positive integers (cf. L. J. Mordell, Diophantine equations, p. 270.

Does this equation have another proof such Lucas's assertion, with short and simple methods? Like this paper: Anglin, W. S. "The Square Pyramid Puzzle." Amer. Math. Monthly 97, 120-124, 1990. The square pyramid puzzle

In the paper,Following two question have simple methods to solve it.

There are no positive integers $x$ such $2x^4+1$ is a square.


There is exactly one positive integer $x$,namely $1$, such that $8x^4+1$ is a square?

But How can I find simple methods to solve $(1)$?

  • 4
    $\begingroup$ What are you asking? As far as I can see there are at least 3 questions in your question, so why don't you post them seperately? $\endgroup$ – Toby Mak May 27 '17 at 12:22
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    $\begingroup$ $$\dfrac{y+1}2\cdot\dfrac{y-1}2=6x^4$$ As $\dfrac{y+1}2-\dfrac{y-1}2=1,\left(\dfrac{y+1}2,\dfrac{y-1}2\right)=1$ and they are of opposite parity. If the highest power of prime $p>3$ that divides $6x^4$ is $a,$ $p^{4a}$ divides exactly one of $$\dfrac{y+1}2,\dfrac{y-1}2$$ $\endgroup$ – lab bhattacharjee Jun 1 '17 at 10:06
  • $\begingroup$ I spent a little bit of time on this and while some of the arguments in the linked paper apply here, I couldn't combine them into a similar cohesive argument (that's not to say such an argument doesn't exist, I just couldn't make one). As a side note, I checked that there are no positive integer solutions to your equation besides $(1,5)$ for $x \leq 300,000$. $\endgroup$ – M10687 Jun 6 '17 at 18:59

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