7
$\begingroup$

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 $x$.

$y^2-1=(y-1)(y+1)=8x^4$

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 ?

$\endgroup$
5
$\begingroup$

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 $x=0$.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.