How can I prove that

$$x^4 - 4y^4 = 2z^2$$

has no solution for positive integers?


  • 1
    $\begingroup$ My failed attempt at modbashing: If $x$ is odd, then the equation fails modulo $2$. Thus $x$ must be even. But then $2^4$ divides the LHS, so $2^4$ must divide the RHS. In particular, $2^3$ must divide $z^2$, so $2^4$ must divide $z^2$. But this means $2^5$ divides the RHS, so $2^5$ must divide the LHS. Now I'm stuck. $\endgroup$ – Zubin Mukerjee Feb 4 '15 at 20:43

Look at the power of two dividing each of the terms.

  • If $2^m$ is the highest power of two dividing $x^4$ then $m$ is necessarily a multiple of four.
  • If $2^n$ is the highest power of two dividing $4y^4$ then $n$ is congruent to two modulo four.
  • If $2^k$ is the hightest power of two dividing $2z^2$ then $k$ is odd.

Hence all those powers of two are distinct. So if we divide the equation by the lowest of them, in the resulting equation there is a single odd term. Contradiction.

So just a $2$-adic argument.

  • $\begingroup$ Thanks a lot I did't know that method, I was struggling with that trying to find a solution which converts the equation into a fermat equation, I liked this approach, thanks a lot $\endgroup$ – Javier Feb 4 '15 at 21:00
  • $\begingroup$ Thanks, @Javier. If you ever have a chance to learn about $p$-adic numbers, then this kind of an argument comes naturally. The idea is used repeatedly there. $\endgroup$ – Jyrki Lahtonen Feb 4 '15 at 21:02

Hint: We show that $|a^4-4b^4|=2c^2$ has no non-zero solutions. The proof is by descent. Suppose there is a solution. Then $a$ and $c$ must be even, say $a=2d$ and $c=2e$. Expand and cancel. We get $|4d^4-b^4|=2e^2$.

Remark: One can alternately rephrase the solution by using the Least Number Principle (well-ordering of the positive integers, aka induction). If there is a non-zero solution of $|a^4-4b^4|=2c^2$, there is a solution with $a^4+4b^4+2c^2$ minimal. But $4d^4+b^4+2e^2\lt a^4+4b^4+2c^2$.

  • $\begingroup$ What is a proof by descent? I don't know the term. EDIT: When you write $a=2c$, is that a typo? $\endgroup$ – Zubin Mukerjee Feb 4 '15 at 20:45
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    $\begingroup$ Infinite descent, a term invented by Fermat. For people who don't know the term, I will add a remark. $\endgroup$ – André Nicolas Feb 4 '15 at 20:57
  • $\begingroup$ @ZubinMukerjee: Thanks, yes it was a typo, the numbers are now called $d$ and $e$, since $c$ was already taken. $\endgroup$ – André Nicolas Feb 4 '15 at 21:00
  • $\begingroup$ Ok I read the new version and I agree, thanks for the solution $\endgroup$ – Javier Feb 4 '15 at 21:03
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    $\begingroup$ You are welcome. I did it this way because it feels Fermat-like, an he used descent. $\endgroup$ – André Nicolas Feb 4 '15 at 21:21

You can relate this to an EC by dividing both of side with y^4. Then we have $2Y^2=X^4-4$ where $X=x/y$ and $Y=z/y^2$. Now you can discuss on the solutions of this curve.


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