On x^2 + y^2 = z^4, x + y = w^2

In a letter to St. Martin and Frenicle on May 31, 1643, Fermat
suggested the problem of finding a Pythagorean triple x,y,z such 
that x+y and x were both squares.  (Actually, he asked for a
rational right triangle whose hypotenuse and the sum of whose 
legs are squares.)  He then gave the solution

   (x,y,z) = (1061652293520, 4565486027761, 4687298610289)

and stated that it was the smallest possible.

Dickson describes several methods, including the one used by Fermat and
another method by Euler, who noted that the problem is equivalent to 
finding two integers x,y such that  x + y = M^2  and  x^2 + y^2 = N^4.
He made x^2 + y^2 a square by setting

              x = p^2 - q^2           y = 2pq

We can make  p^2 + q^2  a square by setting p = r^2 - s^2 and q = 2rs,
which gives x^2 + y^2 = (r^2 + s^2)^4.  This also gives

             x + y = r^4 + 4r^3s - 6r^2s^2 - 4rs^3 + s^4

Euler noticed that if we set s=2 and r=3+v, then

             x + y  =  1 + 148v + 102v^2 + 20v^3 + v^4  

which is  (1 + 74v - v^2)^2  if v = 1343/42.  Therefore, p=1385*1553 
and  q=168*1469, which gives Fermat's solution cited above.