Solution succincta et elegans problematis, quo quaeruntur tres numeri tales, ut tam summae quam differentiae binorum sint quadrata
The problem, according to Euler: "Let x, y and z be the three numbers being sought, of which the largest is x and the smallest z, and let x=pp+qq and y=2pq, so that x+y=(p+q)2 and x–y = (p–q)2. In the same way, setting x=rr+ss and z=2rs, then x+z=(r+s)2 and x–z=(r–s)2. In addition to these four conditions being satisfied, it must be that rr+ss = pp+qq. Then, two additional conditions must be added, that y+z = 2pq+2rs and y–z = 2pq–2rs must both be squares." Euler gets x=50, y=50, z=14, then x=733025, y=488000, z=418304. Then, characteristically, he proposes a slightly different problem (section 16) and solves it by the same means.
Original Source Citation
Mémoires de l'académie des sciences de St.-Petersbourg, Volume 6, pp. 54-65.
Opera Omnia Citation
Series 1, Volume 5, pp.20-27.