# ${x}^{4}-{y}^{4}={z}^{2}$ 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 $d=\mathrm{gcd}(x,y,z)$. 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 |