Editing Proof by infinite descent (section)

===Non-solvability of ''r''<sup>2</sup> + ''s''<sup>4</sup> = ''t''<sup>4</sup> and its permutations===
{{See also|Fermat's right triangle theorem#Fermat's proof}}

The non-solvability of <math>r^2 + s^4 =t^4</math> in integers is sufficient to show the non-solvability of <math>q^4 + s^4 =t^4</math> in integers, which is a special case of [[Fermat's Last Theorem]], and the historical proofs of the latter proceeded by more broadly proving the former using infinite descent. The following more recent proof demonstrates both of these impossibilities by proving still more broadly that a [[Pythagorean triangle]] cannot have any two of its sides each either a square or twice a square, since there is no smallest such triangle:<ref>Dolan, Stan, "Fermat's method of ''descente infinie''", ''[[Mathematical Gazette]]'' 95, July 2011, 269–271.</ref>

Suppose there exists such a Pythagorean triangle. Then it can be scaled down to give a primitive (i.e., with no common factors other than 1) Pythagorean triangle with the same property. Primitive Pythagorean triangles' sides can be written as <math>x=2ab,</math> <math>y=a^2-b^2,</math>  <math>z=a^2+b^2</math>, with ''a'' and ''b'' [[coprime|relatively prime]] and with ''a+b'' odd and hence ''y'' and ''z'' both odd. The property that ''y'' and ''z'' are each odd means that neither  ''y'' nor ''z'' can be twice a square. Furthermore, if ''x'' is a square or twice a square, then each of ''a'' and ''b'' is a square or twice a square. There are three cases, depending on which two sides are postulated to each be a square or twice a square:

*'''''y'' and ''z''''': In this case, ''y'' and ''z'' are both squares. But then the right triangle with legs <math>\sqrt{yz}</math> and <math>b^2</math> and hypotenuse <math>a^2</math> also would have integer sides including a square leg (<math>b^2</math>) and a square hypotenuse (<math>a^2</math>), and would have a smaller hypotenuse (<math>a^2</math> compared to <math>z=a^2+b^2</math>).
*'''''z'' and ''x''''': ''z'' is a square. The integer right triangle with legs <math>a</math> and <math>b</math> and hypotenuse <math>\sqrt{z}</math> also would have two sides (<math>a</math> and <math>b</math>) each of which is a square or twice a square, and a smaller hypotenuse (<math>\sqrt{z}</math> compared to {{nowrap|<math>z</math>)}}.
*'''''y'' and ''x''''': ''y'' is a square. The integer right triangle with legs <math>b</math> and <math>\sqrt{y}</math> and hypotenuse <math>a</math> would have two sides (''b'' and ''a'') each of which is a square or twice a square, with a smaller hypotenuse than the original triangle (<math>a</math> compared to <math>z=a^2+b^2</math>).

In any of these cases, one Pythagorean triangle with two sides each of which is a square or twice a square has led to a smaller one, which in turn would lead to a smaller one, etc.; since such a sequence cannot go on infinitely, the original premise that such a triangle exists must be wrong.

This implies that the equations
:<math>r^2 + s^4 = t^4,</math>
:<math>r^4 + s^2 =t^4,</math> and
:<math>r^4 + s^4 =t^2</math>
cannot have non-trivial solutions, since non-trivial solutions would give Pythagorean triangles with two sides being squares.

For other similar proofs by infinite descent for the ''n'' = 4 case of Fermat's Theorem, see the articles by Grant and Perella<ref>Grant, Mike, and Perella, Malcolm, "Descending to the irrational", ''Mathematical Gazette'' 83, July 1999, pp. 263–267.</ref> and Barbara.<ref>Barbara, Roy, "Fermat's last theorem in the case ''n''&nbsp;=&nbsp;4", ''Mathematical Gazette'' 91, July 2007, 260–262.</ref>