example of Fermat’s last theorem

Fermat stated that for any $n>2$ the Diophantine equation $x^n+y^n=z^n$ has no solution in positive integers. For $n=4$ this follows from the following

Theorem 1.

$x^4+y^4=z^2$ has no solution in positive integers.

Proof.

Suppose we had a positive $z$ such that $x^4+y^4=z^2$ holds. We may assume $\gcd (x,y,z)=1$. Then $z$ must be odd, and $x,y$ have opposite parity. Since ${(x^2)}^2+{(y^2)}^2=z^2$ is a primitive Pythagorean triple, we have

$$x^2=2pq,y^2=q^2-p^2,z=p^2+q^2$$(1)

where $p,q\in \mathbb{N}$, are coprime and have opposite parity. Since $y^2+p^2=q^2$ is a primitive Pythagorean triple, we have coprime $s,r\in \mathbb{N}$, of opposite parity satisfying

$$q=r^2+s^2,y=r^2-s^2,p=2rs.$$(2)

From $\gcd (r^2,s^2)=1$ it follows that $\gcd (r^2,r^2+s^2)=1=\gcd (s^2,r^2+s^2)$, which implies $\gcd (rs,r^2+s^2)=1$. Since ${\left(\frac{x}{2}\right)}^2=\frac{pq}{2}=rs(r^2+s^2)$ is a square, each of $r,s,r^2+s^2$ is a square.

Setting $Z^2=q$, $X^2=r$, $Y^2=s$ $q=r^2+s^2$ leads to

$$Z^2=X^4+Y^4$$(3)

where . Thus, equation 3 gives a solution where . Applying the above steps repeatedly would produce an infinite sequence $z>Z>z_2>\mathrm{\dots }$ of positive integers, each of which was the sum of two fourth powers. But there cannot be infinitely many positive integers smaller than a given one; in particular this contradicts to the fact that there must exist a smallest $z$ for which (1) is solvable. So there are no solutions in positive integers for this equation.∎

A consequence of the above theorem is that the area of a right triangle with integer sides is not a square; equivalently, a right triangle with rational sides has an area which is not the square of a rational.