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

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

has no solutions in positive integers.

Proof.

Suppose the equation has a solution in positive integers, and choose a solution that minimizes $x^2+y^2$. Note that $x,y$, and $z$ are pairwise coprime, since otherwise we could divide out by their common divisor to get a smaller solution. Thus

$$z^2+{(y^2)}^2={(x^2)}^2$$

so that $z,y^2,x^2$ form a pythagorean triple. There are thus positive integers $p,q$ of opposite parity (and coprime since $x,y$, and $z$ are) such that $x^2=p^2+q^2$ and either $y^2=p^2-q^2$ or $y^2=2pq$.

Factoring the original equation, we get

$$(x^2-y^2)(x^2+y^2)=z^2$$

If $y^2=p^2-q^2$, then ${(xy)}^2=p^4-q^4$, and clearly , so we have found a solution smaller than the assumed minimal solution.

Assume therefore that $y^2=2pq$. Now, $x^2=p^2+q^2$; 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

$$q=2PQ,p=P^2-Q^2,x=P^2+Q^2$$

Then

$$PQ(P^2-Q^2)=\frac{1}{2}pq=\frac{y^2}{4}$$

is a square; it follows that $P,Q$, and $P^2-Q^2$ are all (nonzero) squares since they are pairwise coprime. Write

$$P=R^2,Q=S^2,P^2-Q^2=T^2$$

for positive integers $R,S,T$. Then $T^2=R^4-S^4$, 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.

Proof.

Suppose $x,y,z$ is a right triangle with $z$ the hypotenuse, 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

$$x=(2pq)d$$

$$y=(p^2-q^2)d$$

$$z=(p^2+q^2)d$$

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

$$\frac{1}{2}xy=pq(p^2-q^2)d^2$$

must be a square, and thus $pq(p^2-q^2)$ must be a square. Since $p$ and $q$ are coprime, it follows that $p$, $q$, and $p^2-q^2$ are all squares, and thus that $p^2-q^2$ is the difference of two fourth powers. But then

$$\frac{\frac{1}{2}xy}{pq{d}^2}=p^2-q^2$$

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

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