Each of a, b, c, and d is an integer that satisfies this system of equations:
ab+cd = 34
ac-bd =19
Determine all possible values of a, b, c, and d.
I started by rearranging the equations into
abc=34-c^2*d and abc=19b+b^2*d
Then subtract the first from the second to get
b^2*d+c^2*d+19b-34c=0
Next step is to multiple through by d, then the equation becomes quadratic in bd and cd:
(bd)^2+(cd)^2+19(bd)-34(cd)=0
This is effectively a circle, so complete the square to get
(2bd+19)^2 + (2cd-34)^2 = 34^2+19^2 = 1517
So we need decompositions of 1517 into squares, there are two: 34^2+19^2 and 26^2+29^2. By parity, the even squares pair with 2cd-34 and the odd squares pair with 2bd+19.
Then there are eight subcases:
2bd+19=19 and 2cd-34=34 -> bd=0 and cd=34
2bd+19=19 and 2cd-34=-34 -> bd=0 and cd=0
2bd+19=-19 and 2cd-34=34 -> bd=-19 and cd=34
2bd+19=-19 and 2cd-34=-34 -> bd=-19 and cd=0
2bd+19=29 and 2cd-34=26 -> bd=5 and cd=30
2bd+19=29 and 2cd-34=-26 -> bd=5 and cd=4
2bd+19=-29 and 2cd-34=26 -> bd=-24 and cd=30
2bd+19=-29 and 2cd-34=-26 -> bd=-24 and cd=4
Solving for each of these eight subcases eventually gets us to these solutions:
bd=0 and cd=34 -> (a b c d) = (-19 0 -1 -34) or (19 0 1 34)
bd=0 and cd=0 -> (a b c d) = (-1 -34 -19 0) or (1 34 19 0)
bd=-19 and cd=34 -> (a b c d) = (0 -19 34 1) or (0 19 -34 -1)
bd=-19 and cd=0 -> (a b c d) = (-34 -1 0 19) or (34 1 0 -19)
bd=5 and cd=30 -> (a b c d) = (-4 -1 -6 -5) or (4 1 6 5)
bd=5 and cd=4 -> (a b c d) = (-6 -5 -4 -1) or (6 5 4 1)
bd=-24 and cd=30 -> (a b c d) = (-1 -4 5 6) or (1 4 -5 -6)
bd=-24 and cd=4 -> (a b c d) = (-5 -6 1 4) or (5 6 -1 -4)
Solution
