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# infinite descent

Fermat invented this method of infinite descent. The idea is: If a given natural number $n$ with certain properties implies that there exists a smaller one with these properties, then there are infinitely many of these, which is impossible.

Here is an example:

Let $a=2mn$, $b=n^{2}-m^{2}$, $c=m^{2}+n^{2}$. Then $\{a,b,c\}$ is a primitive Pythagorean triple, and the area $A$ of the right triangle with sides $a,b,c$ is $ab/2=mn(n^{2}-m^{2})$.

Suppose $A$ is a square. Then, since $m,n$ are coprime and of opposite parity, $\gcd(m+n,m-n)=\gcd(m,n)=1$. Thus, for $A$ to be a square, each of $m,n,m-n,m+n$ must be squares itself. Setting $r^{2}=m$, $s^{2}=n$, we have $A=(rs)^{2}(s^{4}-r^{4})$.

We prove that the Diophantine equation $x^{4}-y^{4}=z^{2}$ has no solution in natural numbers.

###### Remark 1.

Suppose that $z^{2}+y^{4}=x^{4}$, where $\gcd(x,y,z)=1$, $x,y,z\in{\mathbb{N}}$. Then $x$ is odd, and $y,z$ have opposite parity.

###### Proof.

If $x$ was even, then $x^{4}=z^{2}+y^{4}\equiv(z+y^{2})^{2}\equiv 0\;\;(\mathop{{\rm mod}}2)$, so $z,y^{2}\equiv 0\;\;(\mathop{{\rm mod}}2)$ or $z,y^{2}\equiv 1\;\;(\mathop{{\rm mod}}2)$. But $z,y^{2}\equiv 0\;\;(\mathop{{\rm mod}}2)$ conflicts with $\gcd(x,y,z)=1$. And $z,y^{2}\equiv 1\;\;(\mathop{{\rm mod}}2)$ implies $y^{2}+(z^{2})^{2}\equiv 2\;\;(\mathop{{\rm mod}}4)$ contradicting $x^{4}\equiv 0\;\;(\mathop{{\rm mod}}4)$. Thus, $x$ is odd, and $x^{4}=z^{2}+(y^{2})^{2}\equiv(z+y^{2})^{2}\equiv 1\;\;(\mathop{{\rm mod}}2)$ implies that $z,y^{2}$ have opposite parity. ∎

Suppose $x$ is odd and $z$ is even. Then we have $z=2pq$, $y^{2}=q^{2}-p^{2}$ and $x^{2}=q^{2}+p^{2}$, where $p,q$ have opposite parity and are coprime. Since $z$ is odd, this implies $(xy)^{2}=q^{4}-p^{4}$, so it is sufficient to show that there is no solution for odd $z$.

Now $x,z$ are assumed odd. Then $y$ is even, and there exist $m,n\in{\mathbb{N}}$, $m<n$,$(2mn,m+n)=1$ such that

$\displaystyle y^{2}$ | $\displaystyle=2mn$ | (1) | |||

$\displaystyle x^{2}$ | $\displaystyle=n^{2}$ | $\displaystyle+m^{2}$ | (2) | ||

$\displaystyle z$ | $\displaystyle=n^{2}$ | $\displaystyle-m^{2}.$ | (3) |

Since $m^{2}+n^{2}=x^{2}$ is a primitive Pythagorean triple, there exist $p,q\in{\mathbb{N}}$, $p<q$, $(2pq,p+q)=1$ satisfiying

$\displaystyle m$ | $\displaystyle=2pq$ | (4) | |||

$\displaystyle n$ | $\displaystyle=q^{2}$ | $\displaystyle-p^{2}$ | (5) | ||

$\displaystyle x$ | $\displaystyle=q^{2}+p^{2}.$ | (6) |

Since $2mn$ is a square and $m,n$ are coprime and, say, $n$ is odd, $n$ is a square, and we have $m=2r^{2}$, $n=s^{2}$.

From the primitive Pythagorean triple $m^{2}+n^{2}=x^{2}$ we get $x=u^{2}+v^{2}$, $n=u^{2}-v^{2}$, $m=2uv$. Since $2uv=2r^{2}$ $uv$ is a square, and each of $u$ and $v$ is a square: $u=g^{2}$, $v=h^{2}$.

Substituting $n,u,v$ in $n=u^{2}-v^{2}$ we have $s^{2}=g^{4}-h^{4}$. But since $n+(1/2)<m$ this implies $n=s^{2}<z=n^{2}-m^{2}<z^{2}$, thus we have another solution with odd $s<z$. This contradicts to the fact that there exists a smallest solution.

## Mathematics Subject Classification

11D25*no label found*

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