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# $x^{4}-y^{4}=z^{2}$ 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 $p^{2}+q^{2}=x^{2}<x^{2}+y^{2}$, 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

$q=2PQ,\quad p=P^{2}-Q^{2},\quad x=P^{2}+Q^{2}$ |

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

$P=R^{2},\quad Q=S^{2},\quad P^{2}-Q^{2}=T^{2}$ |

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

$R^{2}+S^{2}=P+Q<(P+Q)(PQ)(P-Q)=\frac{1}{2}pq=\frac{y^{2}}{4}\leq y^{2}<x^{2}+y% ^{2}$ |

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

$\displaystyle x=(2pq)d$ | ||

$\displaystyle y=(p^{2}-q^{2})d$ | ||

$\displaystyle z=(p^{2}+q^{2})d$ |

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

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

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

$\frac{\frac{1}{2}xy}{pqd^{2}}=p^{2}-q^{2}$ |

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

## Mathematics Subject Classification

11D41*no label found*14H52

*no label found*11F80

*no label found*

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