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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 $m,n$ be coprime positive integers with opposite parity, $m<n$ , and, say, $m$ is even.
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 $\gcd(x,y,z)=1$ , $x,y,z \in \N$ . Then $x$ is odd, and $y,z$ have opposite parity.
Proof. If $x$ was even, then $x^4=z^2+y^4 \equiv (z+y^2)^2 \equiv 0 \pmod{2}$ , so $z, y^2 \equiv 0 \pmod{2}$ or $z, y^2 \equiv 1 \pmod{2}$ . But $z, y^2 \equiv 0 \pmod{2}$ conflicts with $\gcd(x,y,z)=1$ . And $z,y^2 \equiv 1 \pmod 2$ implies $y^2+(z^2)^2\equiv 2 \pmod{4}$ contradicting $x^4 \equiv 0 \pmod{4}$ . Thus, $x$ is odd, and $x^4=z^2+(y^2)^2 \equiv (z+y^2)^2 \equiv 1 \pmod{2}$ 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 $m,n \in \N$ , $m<n$ ,$(2mn,m+n)=1$ such that \begin{eqnarray} \label{eq1} y^2&=2mn \\ x^2&=n^2&+m^2 \\ z&=n^2&-m^2. \end{eqnarray}Since $m^2+n^2=x^2$ is a primitive Pythagorean triple, there exist $p,q \in \N$ , $p<q$ , $(2pq,p+q)=1$ satisfiying \begin{eqnarray} \label{eq2} m&=2pq \\ n&=q^2&-p^2 \\ x&=q^2+p^2. \end{eqnarray}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.
See here for a discussion of infinite descent vs. induction.
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