Fermat's Last Theorem for n = 4

The accepted point of view on the possible progress on what is nowadays call Fermat's Last Theorem by Fermat himself is that most likely he had a proof of the impossibility of the solution of

x^n + y^n = z^n

in integers for n = 4.  The statement is easily proved by the method of Infinite Descent - which was rightfully claimed by Fermat as his own discovery. So it is quite probable that the proof below was indeed known to Fermat himself.

First of all, if there is a solution x, y, z,   there is a primitive solution, i.e. such a triple that gcd(x, y, z) = 1.  This means that the three numbers are pairwise relatively prime.

Second, we replace x^4 + y^4 = z^4  with x^4 + y^4 = z^2,  for if the latter has no integer solutions, the same holds for the former due to x^4 + y^4 = (z^2)^2.

So, the task is to prove that x^4 + y^4 = z^2  has no integer solutions. Suppose on the contrary that for some x, y, z,  with gcd(x, y, z) = 1,   x^4 + y^4 = z^2.  Then, of course, (x^2)^2 + (y^2)^2 = z^2  so that x^2, y^2, z  is a Pythagorean triple which we may assume primitive. Hence (if necessary after swaping x   and y )  there are mutually prime p, q,  of opposite parity, with p \gt q \gt 0  and

1. x^2 = 2pq,
2. y^2 = p^2 - q^2,
3. z = p^2 + q^2.

It follows that there are positive, mutually prime a, b  of opposite parity such that a \gt b \gt 0  and

1. q = 2ab,
2. y = a^2 - b^2,
3. p = a^2 + b^2.

Substituting we get

x^2 = 2pq = 4ab(a^2 + b^2).

We are going to show that all three a, b, a^2 + b^2  are squares.

First, since x  is even, ab(a^2 + b^2 = (x/2)^2 is a square of an integer. If a prime s  divides ab   then, since they are mutually prime, s   divides exactly one of a  or b.  Let, s|a.  Then, of course, s  cannot divide a^2 + b^2  ,hence gcd(ab, a^2 + b^2) = 1 . Thus ab  and a^2 + b^2  are squares, and since gcd(a, b) = 1,   so are a  and b.

Let a = X^2, b = Y^2, a^2 + b^2 = Z^2.  Then X^4 + Y^4 = Z^2.   It appears that the existence of a primitive triple x, y, z  led to the existence of another primitive triple X, Y, Z.  But

Z^2 = X^4 + Y^4 = a^2 + b^2 = p \lt p^2 + q^2 = z \lt z^2.

Which serves the starting point to the infinite descent. A contradiction.