How do I prove that this Diophantine equation has no solutions?

Question

How can I prove that

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

has no solution for positive integers?

Thanks.

Answer 1

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.

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So just a $2$-adic argument.

Answer 2

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 $4d^4+b^4+2e^2\lt a^4+4b^4+2c^2$.

Answer 3

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.