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$x^4-y^4=z^2$ has no solutions in positive integers (Theorem)

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  
$\displaystyle x^4-y^4=z^2$
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. Then
$\displaystyle (x^2-y^2)(x^2+y^2)=z^2$
and also $ x^2-y^2$ and $ x^2+y^2$ are coprime. Thus $ x^2+y^2=r^2$, $ x^2-y^2=s^2$. This shows that $ x$ and $ y$ must be of opposite parity (they cannot both be even, and if both were odd, then one of $ x^2+y^2$ and $ x^2-y^2$ would be $ \equiv 3\pmod 4$). Now, let
$\displaystyle u=\frac{r-s}{2}$    
$\displaystyle v=\frac{r+s}{2}$    

Then
$\displaystyle \frac{1}{2}uv=\frac{y^2}{4}$
and thus $ \frac{1}{2}uv$ is a square, and also $ y$ is even, hence $ x$ is odd.

Now, note that $ u^2+v^2=x^2$, so exactly one of $ u,v$ is even. By relabeling if necessary, assume that $ u$ is even. Then we may write $ u=2PQ, v=P^2-Q^2, x=P^2+Q^2$. But then $ \frac{1}{2}uv=PQ(P^2-Q^2)$ is a square. It follows that $ P,Q$, and $ P^2-Q^2$ are all squares. Writing, then, $ P=R^2, Q=S^2, P^2-Q^2=T^2$, we have $ T^2=R^4-S^4$. But

$\displaystyle R^2+S^2=P+Q\leq (P+Q)(PQ)(P-Q)=\frac{1}{2}uv=\frac{y^2}{4}\leq y^2< x^2+y^2$
where the strict inequality follows since we are assuming nonzero integers.

Finally, we must show that $ R,S>0$. Note that $ r\neq s$ since $ y\neq 0$, and that $ r\neq 0$. Thus $ u>0$ and $ v>0$. But then $ P,Q>0$ and $ P\neq Q$. Finally, $ R,S>0$ since $ P,Q>0$.

We have thus found a smaller solution in positive integers, contradicting the hypothesis. $ \qedsymbol$

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
$\displaystyle x=(2pq)d$    
$\displaystyle y=(p^2-q^2)d$    
$\displaystyle z=(p^2+q^2)d$    

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

$\displaystyle \frac{1}{2}xy=pq(p^2-q^2)d^2$
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
$\displaystyle \frac{\frac{1}{2}xy}{pqd^2}=p^2-q^2$
must also be a square. Since both $ p$ and $ q$ are squares, this is impossible by the theorem. $ \qedsymbol$



"$x^4-y^4=z^2$ has no solutions in positive integers" is owned by rm50.
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See Also: example of Fermat's last theorem
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Cross-references: triangle's, hypotenuse, hypothesis, strict inequality, necessary, odd, even, parity, opposite, coprime, divide, pairwise coprime, equation, integers, positive, solutions, infinite descent, sides, integral, right triangle, area, difference, square, sum, example of Fermat's last theorem
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This is version 2 of $x^4-y^4=z^2$ has no solutions in positive integers, born on 2007-05-13, modified 2007-05-15.
Object id is 9377, canonical name is X4Y4z2HasNoSolutionsInPositiveIntegers.
Accessed 726 times total.

Classification:
AMS MSC: 11D41 (Number theory :: Diophantine equations :: Higher degree equations; Fermat's equation)
 14H52 (Algebraic geometry :: Curves :: Elliptic curves)
 11F80 (Number theory :: Discontinuous groups and automorphic forms :: Galois representations)
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