If a, b are integers, and (a−b)^2 + 8b^2 = 108, what is the number of the ordered pairs (a,b)?

Question

Suraj Pandey · Accepted Answer

Understand the problemWe need to find integer solutions for the equation (a&minus;b)^2 + 8b^2 = 108.
Substitute different values of b and solve for aLet k = a - b. Then the equation becomes k^2 + 8b^2 = 108.
Test values of bb = 0: k^2 + 0 = 108 -> No solution (k^2 = 108 is not a perfect square)b = 1: k^2 + 8 = 108 -> k^2 = 100 -> k = &plusmn;10 -> (a, b) = (11, 1), (-9, 1)b = 2: k^2 + 32 = 108 -> k^2 = 76 -> No solution (k^2 = 76 is not a perfect square)b = 3: k^2 + 72 = 108 -> k^2 = 36 -> k = &plusmn;6 -> (a, b) = (9, 3), (-3, 3)b = 4: k^2 + 128 = 108 -> No solution (k^2 = -20 is not possible)b = -1: k^2 + 8 = 108 -> k^2 = 100 -> k = &plusmn;10 -> (a, b) = (9, -1), (-11, -1)b = -3: k^2 + 72 = 108 -> k^2 = 36 -> k = &plusmn;6 -> (a, b) = (3, -3), (-9, -3)
Count the solutionsThe ordered pairs (a, b) are:(11, 1), (-9, 1), (9, 3), (-3, 3), (9, -1), (-11, -1), (3, -3), (-9, -3)
ConclusionThere are 8 ordered pairs (a, b).