Quartic equation $x^4-y^4=2$ over rationals

Question

Was it already proved that $x^4-y^4=2$ has no solutions over rationals?

Or for simpler sub-problems $x^2-y^4=2$ or $x^4-y^2=2$.

Answer 1

We have the following elementary proof there's no solution. Let's rewrite the equation $$ (x - y)(x + y)(x^2 + y^2) = x^4 - y^4 = 2z^4 $$ with $ x, y, z \in \mathbb Z $ and $ \gcd(x, y, z) = 1 $.

For all prime $ p \ne 2 $, we have $ 4 \mid 4v_p(z) = v_p(x - y) + v_p(x + y) + v_p(x^2 + y^2) $. As the $ \gcd $ of any two of $ x - y, x + y, x^2 + y^2 $ is at most $ 2 $ (well known result from cyclotomic polynomials, or basic $ \gcd $ manipulations), we also have that at most one of $ v_p(x - y), v_p(x + y), v_p(x^2 + y^2) $ is nonzero.

So $ 4 \mid v_p(x - y), v_p(x + y), v_p(x^2 + y^2) $, which implies that $ x - y, x + y, x^2 + y^2 $ are all three a $ 4 $-th power times a power of $ 2 $ (which we can assume to be one of $ 1, 2, 4, 8 $ by "absorbing" higher powers in the $ 4 $-th power).

If one of $ x $ or $ y $ is even, then the other is too and $ 16 \mid 2z^4 $ which means that $ 2 \mid z, \gcd(x, y, z) $. Absurd. So $ x $ and $ y $ are both odd and $ v_2(x^2 + y^2) = 1 $.

We'll use the fact that $ 2(x^2 + y^2) = (x - y)^2 + (x + y)^2 $. We distinguish cases depending on how the $ 2 $s are distributed among the three factors:

• $ x - y = a^4, x + y = b^4, x^2 + y^2 = 2c^4 $ which gives $ (a^2)^4 + (b^2)^4 = (2c^2)^2 $.
• $ x - y = 2a^4, x + y = 8b^4, x^2 + y^2 = 2c^4 $ which gives $ (a^2)^4 + (2b^2)^4 = (c^2)^2 $.
• $ x - y = 4a^4, x + y = 4b^4 $ implies that $ x = 2a^4 + 2b^4 $ is even (as well as $ y $). Absurd.
• $ x - y = 8a^4, x + y = 2b^4 $ is the same as the second case but replacing $ y $ by $ -y $.

But $ a^4 + b^4 = c^2 $ is known to have no solutions in integers. One can prove this using the general form of a Pythagorean triple to derive an infinite descent.