How can I prove that $y^2=8x^4+1$ has no integral solution with $x\ge 2$ with elementary methods ?

With elementary I mean using only modular arithmetic, the unique factorization theorem and the theory of quadratic residues modulo a prime $p$.

I tried various approaches , but I did not manage to prove the claim :

First of all, if $x\ge 2$, there is a prime factor $p$ dividing $x$.


So, if $y-1$ or $y+1$ is divisible by an odd prime, we have $p^4|y-1$ respective $p^4|y+1$. Does this lead to anywhere ?

$y^2-9=8x^4-8$ , which implies $(y-3)(y+3)=8(x-1)(x+1)(x^2+1)$. Can I make use of the fact that every odd prime $p$ dividing $x^2+1$ has the form $4k+1$ ?

Since the equation is closely related to the triangular numbers (The question whether a triangular number can be a fourth power leads to the given equation), I also studied the convergents of $\sqrt{2}$ (which satisfy the pell-equation $x^2-2y^2=1$).

The sequence $B_n$ of the positive integers, whose squares are triangular satisfy the recurrence relation $B_1=1$ , $B_2=6$ , $B_n=6B_{n-1}-B_{n-2}$. So, showing that no $B_n$ except $1$ is a perfect square would also finish the proof.

Any ideas ?


The positive integer $y$ must be odd, and letting $z=(y-1)/2$ we get $z(z+1)=2x^4.$ The two consecutive integers $z$ and $z+1$ have no common prime factors, so one is a fourth power and the other twice a fourth power.

Case 1: If $z=\ell^4$ and $z+1=2m^4$, then $\ell^4+1=2m^4$. From my answer here, we see that $\ell=1$ and so $y=3$ and $x=1$.

(This part uses the fact that $z^2=x^4-y^4$ has no solutions in non-zero integers. This is Exercise 1.6 in Edwards's book on Fermat's Last Theorem. The proof uses the representation of Pythagorean triples and infinite descent.)

Case 2: If $z=2m^4$ and $z+1=\ell^4$, then $\ell^4-1=2m^4$. Since $(\ell^2-1)(\ell^2+1)=2m^4$ and $\gcd(\ell^2-1,\ell^2+1)\leq 2$, one of the factors $\ell^2-1$ or $\ell^2+1$ is a fourth power, in particular a square. The only two consecutive squares are $0$ and $1$, so we must have $\ell=1$. This implies $z=0$ and so $y=1$ and $x=0$.

  • $\begingroup$ With infinite descent, it should be possible to show that $x^2=y^4-z^4$ has no non-zero integer solution, so this would complete the elementary proof. If you add this detail and show how we can use it to solve case $1$, I will accept the answer. $\endgroup$ – Peter Mar 16 '17 at 23:13
  • $\begingroup$ I will try the infinite descent myself. What was the method to arrive at $x^2=y^4-z^4$ ? Squaring and subtracting a fourth power ? $\endgroup$ – Peter Mar 16 '17 at 23:21
  • $\begingroup$ @Peter Click on the link that says "From my answer here" and you will see the argument. $\endgroup$ – user940 Mar 16 '17 at 23:23
  • $\begingroup$ OK, I have done this. Only $4$ lines. It would have been nice if you would just have posted them ... So, I will go on with the infinite descent to complete the proof. $\endgroup$ – Peter Mar 16 '17 at 23:28
  • 1
    $\begingroup$ @Peter Glad to help out! $\endgroup$ – user940 Mar 17 '17 at 2:21

Here is another approach. We can rewrite original equation as

$y^2 - 1 = 2^kx^4$ where $x$ is odd. Equation (1)

$(y-1)(y+1) = 2^k m^4 n^4$ where $(m,n) = 1$ and both $m$ and $n$ are odd.

$gcd(y-1, y+1) = 2$

STEP 1: Without loss of generality, we can write $y - 1 = 2^{k-1} m^4$ and $y + 1 = 2n^4$. The difference gives us,

$n^4 - 2^{k-2} m^4 = 1$

We have an equation that looks like Equation(1) with $n^2 < y, m < x$ and so this descent will lead us to an equation of the form $r^4 - 2s^4 = 1$ Equation (2)

or $r^4 - s^4 = 1$ Equation (3)

with gcd(r,s) = 1 and $r, s$ being odd

The LHS and RHS of Equation (2) (and Equation (3)) are not congruent mod 4.

Hence Equations (2) and (3) have no solution besides the trivial solution $r = 1, s = 0$

Hence Equation (1) has no solution in integers besides $x = \pm1$

ps: If in STEP 1, we interchanged the values of $y-1$ and $y+1$, we will get

$2^{k-2}m^4 - n^4 = 1 \implies 2^{k-2}m^4 = (n^2)^2 + 1^2$

Since $k$ is even, this is of the form $p^2 = n^4 + 1^4$ which do not have solutions in integer $p, n$ per Fermat.

  • $\begingroup$ In your PS, you write that $k$ is even. In fact, $k$ is odd. $\endgroup$ – user940 Mar 18 '17 at 14:19
  • $\begingroup$ Agree. Separate question. Isn't the original problem same as Mordell's Diophantine equation? $y^2 = D x^4 + 1$ $\endgroup$ – sku Mar 19 '17 at 2:26

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