| No Solution Yet | Submitted by Danish Ahmed Khan |
| Rating: 5.0000 (1 votes) |
Solution
|
 | Comment 3 of 4 |  |
Add 1 to A, B, and C. Then each becomes a perfect square:
A+1 = x^4+8x^3+26x^2+40x+24 = (x^2+4x+5)^2
B+1 = y^4+4y^3+8y^2+8y+3 = (y^2+2y+2)^2
C+1 = x^4+8x^3+30x^2+56x+48 = (x^2+4x+7)^2
Then A+1, B+1, and C+1 are ascending perfect squares. But there is only one square between A+1 and C+1. Therefore B+1 must equal (x^2+4x+6)^2.
Then (y^2+2y+2)^2 = (x^2+4x+6)^2. Taking square roots leaves two cases: y^2+2y+2 = -(x^2+4x+6) and y^2+2y+2 = x^2+4x+6.
y^2+2y+2 = -(x^2+4x+6) can be rearranged into (y+1)^2 + (x+2)^2 + 3 = 0. The left side is strictly positive so there are no solutions in this case.
y^2+2y+2 = x^2+4x+6 can be rearranged into (y+1)^2 =(x+2)^2 + 1. This implies (y+1)^2 and (x+2)^2 are squares differing by 1; the only such pair of squares is (0,1). Then (y+1)^2=1 and (x+2)^2=0. This yields two solutions (x,y) = (-2,0) or (-2,-2).
| Posted by Brian Smith on 2020-09-23 10:31:16 |
