CHAP. XVIII] FOUR QUADKATIC FUNCTIONS MADE SQUARES. 495
The final expression is a square if and only if
P<?(P4~"#4) -^s(r4— s4) = D. By special assumptions, Euler was led to the values
Then
The condition is thus (a +3/3) (a +7/3) (a2 +/32) = D, which is treated by the usual methods. From a = 2, j8 = 1 and a= — 17, /-. =7, we get the solutions
p = iL??! 5-292 . 132-532 37-132-532
' 8-92 ' 32-112' 3-4-7-592' 3-7-42.52-192* Euler20 made x+y±x2, x+y^y* all squares. Replace rr, y by rc/0, y/«. Then (x+y*)z±x*, (x+y)z±y* are to be squares. This will be the case if
xz = 2AB, 2/2 = 2CD, (x+y)z = A2+52 = C2+D2. The final equality holds if
A=ac+H B = ad-bc, C=ad+6c, D = ac~-bd.
The first two conditions hold ttx=Af, y^Cg, 2B=Af~, 2D = Cg2. By the
latter,
which are equal if
Hence the problem will be solved if we make R = \3. Set g — \. Since R = Q f or / = 1 (which makes x = y),-we take / = 1 + $• Then 12 is the square of 5+2£+13Z2/5 if £ = 60/11. Dropping the common factor 13 in x, y, we get
C 072 «12
* = 4- 11-71, 0-4-37.61, «- 2 49.81 -
» M«m. Acad. So. St. PStersbourg, 11, 1830 (1780), 46; Comm. Arith., II, 447.