Math: Formula for Squares

Formula for Squares

Show that if x is a postive integer, and both 2x + 1 and 3x + 1 are squares, then x is a multiple of 40.

Solution

We work with the modulo classes of 5 and 8. Any square number must leave remainder 0, 1 or 4 when divided by 5 or 8:

n (mod5) n2 (mod 5) 0 0 1 1 2 4 3 4 4 1

n (mod8) n2 (mod 8) 0 0 1 1 2 4 3 1 4 0 5 1 6 4 7 1

Now lets examine the remainders when 2x + 1 and 3x + 1 are divided by 5 and 8:

x (mod5) 2x+1 (mod 5) 3x+1 (mod 5) 0 1 1 1 3 4 2 0 2 3 2 0 4 4 3

x (mod8) 2x+1 (mod 8) 3x+1 (mod 8) 0 1 1 1 3 4 2 5 7 3 7 2 4 1 5 5 3 0 6 5 3 7 7 6

From the table on the left, we see that it is only when x = 0 (mod 5) that both 2x + 1 and 3x + 1 have remainders 0, 1 or 4 mod 5 that allow both of them to be squares. Similarly from the second table, we must have x = 0 (mod 8).

Thus 5 and 8 both divide x, and so x is a multiple of 40.

The first few values of x with both 2x + 1 and 3x + 1 square are x = 0, 40, 3960, 388080, 38027920, 3726348120, ...