Solve the Diophantine Equation $x^2 + 1 = 2y^4$ over $\mathbb{Z}$.

Solve the Diophantine Equation $x^2 + 1 = 2y^4$ over $\mathbb{Z}$.

I have found few elementary solutions like $(1,1)$.

I have tried it with variable replacements. After solving it a bit it becomes clear that both $x$ and $y$ are odd. Evaluating it for $x = 2x_1+1$ and $y=2y_1+1$ I have reached:

$$2x_1^2 + 2x_1 + 1 = (2y_1 + 1)^4$$

or

$$x_1^2 + (x_1 + 1)^2 = (2y_1 + 1)^4$$

I don't know how to solve this further. Do you have any ideas, hints or techniques to solve this?

Thanks.

• @Jean Corrected it. – TheRandomGuy Mar 28 '16 at 11:14
• I know the Pythagoras Theorem but what do I do with it. Also, the hypotenuse must be a perfect square. – TheRandomGuy Mar 28 '16 at 11:19
• So $(\pm 239,\pm 13),(\pm 1, \pm 1)$ appears to be the only solutions. – S.C.B. Mar 28 '16 at 11:35
• @K.K.McDonald Given the reputation of Smarandache, the article has to be carefully checked I guess :) – Jean-Claude Arbaut Mar 28 '16 at 11:38
• xD ok I googled him and now I know what you mean! – K.K.McDonald Mar 28 '16 at 11:39

The only solutions to this equation are with $(x,y)=(1,1)$ and $(239,13)$. This was first proved by Ljunggren via an elaborate version of Skolem's $p$-adic method. Other proofs all rely upon pretty sophisticated techniques (linear forms in logarithms, the hypergeometric method, Frey curves, etc). Mordell asked many years ago (as reported in Guy's UPINT) for an elementary proof. As of now, there is none known.