x4-y4=z2 has no solutions in positive integers

We know (see example of Fermat’s Last Theorem) that the sum of two fourth powers can never be a square unless all are zero. This article shows that the difference of two fourth powers can never be a square unless at least one of the numbers is zero. Fermat proved this fact as part of his proof that the area of a right triangleMathworldPlanetmath with integral sides is never a square; see the corollary below. The proof of the main theorem is a great example of the method of infinite descent.

Theorem 1.

has no solutions in positive integers.


Suppose the equation has a solution in positive integers, and choose a solution that minimizes x2+y2. Note that x,y, and z are pairwise coprime, since otherwise we could divide out by their common divisorMathworldPlanetmathPlanetmath to get a smaller solution. Thus


so that z,y2,x2 form a pythagorean tripleMathworldPlanetmath. There are thus positive integers p,q of opposite parity (and coprimeMathworldPlanetmath since x,y, and z are) such that x2=p2+q2 and either y2=p2-q2 or y2=2pq.

Factoring the original equation, we get


If y2=p2-q2, then (xy)2=p4-q4, and clearly p2+q2=x2<x2+y2, so we have found a solution smaller than the assumed minimal solution.

Assume therefore that y2=2pq. Now, x2=p2+q2; we may assume by relabeling if necessary that q is even and p odd. Then p,q,x are pairwise coprime and form a pythagorean triple; thus there are P>Q>0 of opposite parity and coprime such that




is a square; it follows that P,Q, and P2-Q2 are all (nonzero) squares since they are pairwise coprime. Write


for positive integers R,S,T. Then T2=R4-S4, and


We have thus found a smaller solution in positive integers, contradicting the hypothesis. ∎

Corollary 1.

No right triangle with integral sides has area that is an integral square.


Suppose x,y,z is a right triangle with z the hypotenuseMathworldPlanetmath, and let d=gcd(x,y,z). Either x/d or y/d is even; by relabeling if necessary, assume x/d is even. Then we can choose relatively prime integers p,q with p>q and of opposite parity such that


If the triangle’s area is to be a square, then


must be a square, and thus pq(p2-q2) must be a square. Since p and q are coprime, it follows that p, q, and p2-q2 are all squares, and thus that p2-q2 is the difference of two fourth powers. But then


must also be a square. Since both p and q are squares, this is impossible by the theorem. ∎

Title x4-y4=z2 has no solutions in positive integers
Canonical name X4y4z2HasNoSolutionsInPositiveIntegers
Date of creation 2013-03-22 17:05:04
Last modified on 2013-03-22 17:05:04
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 7
Author rm50 (10146)
Entry type Theorem
Classification msc 11D41
Classification msc 14H52
Classification msc 11F80
Related topic ExampleOfFermatsLastTheorem
Related topic IncircleRadiusDeterminedByPythagoreanTriple