How can I prove that
$$x^4 - 4y^4 = 2z^2$$
has no solution for positive integers?
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.Sign up to join this community
Look at the power of two dividing each of the terms.
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.
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 $4d^4+b^4+2e^2\lt a^4+4b^4+2c^2$.