(2u)² – 3 = 4u² – 3 = 4(x² + x + 1) – 3 = 4x² + 4x + 1 = (2x + 1)².

This shows that (2u)² – 3 is a perfect square.

Hence

(2u)² – 3 = (2x + 1)², or (2u)² – (2x + 1)²= 3.

Factorising the difference of two squares, we get

(2u + 2x + 1)(2u – 2x – 1) = 3

3 can only be factorized as 1 x 3 or -1 x -3, so we have just four cases to consider.

Case 1

2u + 2x + 1 = 3 & 2u – 2x – 1 = 1.

Adding these equations we get 4u = 4, so u = 1 and x = 0.

Case 2

2u + 2x + 1 = 1 & 2u – 2x – 1 = 3.

Adding these equations, we again get 4u = 4, so u = 1 and x = -1 .

Case 3

2u + 2x + 1 = - 3 & 2u – 2x – 1 = - 1.

Adding these equations, we get 4u = -4, so u = -1 and x = -1 .

Case 4

2u + 2x + 1 = - 1 & 2u – 2x – 1 = - 3.

Adding these equations, we again get 4u = -4, so u = -1 and x = 0 .

Hence the only integer values of x which give an integer value for u are x = -1 and 0.

Challenge 317: Is u an integer?