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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.

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    $\begingroup$ @Jean Corrected it. $\endgroup$ Mar 28, 2016 at 11:14
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    $\begingroup$ I know the Pythagoras Theorem but what do I do with it. Also, the hypotenuse must be a perfect square. $\endgroup$ Mar 28, 2016 at 11:19
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    $\begingroup$ So $(\pm 239,\pm 13),(\pm 1, \pm 1)$ appears to be the only solutions. $\endgroup$
    – S.C.B.
    Mar 28, 2016 at 11:35
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    $\begingroup$ @K.K.McDonald Given the reputation of Smarandache, the article has to be carefully checked I guess :) $\endgroup$ Mar 28, 2016 at 11:38
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    $\begingroup$ xD ok I googled him and now I know what you mean! $\endgroup$ Mar 28, 2016 at 11:39

2 Answers 2

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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.

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If $x$ and $y$ are integers such that $x^2 + 1 = 2y^4$ then in $\Bbb{Z}[i]$ we have $$2y^4=(x+i)(x-i).$$ Of course $x$ is odd and so the greatest common divisor of the two factors on the right hand side is $1+i$. Because $\Bbb{Z}[i]$ is a unique factorization domain it follows that $$x+i=i^k(1+i)(u+vi)^4,$$ for some integers $k$, $u$ and $v$. Expanding the right hand side shows that $$x+i=i^k((u^4-4u^3v-6u^2v^2+4uv^3+v^4)+(u^4+4u^3v-6u^2v^2-4uv^3+v^4)i),$$ and so comparing imaginary parts and changing the sign of $u$ if necessary, we see that $$u^4+4u^3v-6u^2v^2-4uv^3+v^4=\pm1.$$ This is a Thue equation, which has only finitely many integral solutions and there exist effective methods to find them all. With the help of a computer I've found that they are, up to sign, $$(0, 1),\quad(1, 0),\quad (2, -3),\quad (3, 2).$$ These correspond to $x=\pm1$ and $x=\pm239$, with $y=\pm1$ and $y=\pm13$ respectively.

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