Perfect Square

Rewriting:

y = 8x² + 1

Let y = n² where n is a positive integer, then

n² = 8x² + 1
n² - 1 = 8x²
(n + 1)(n - 1) = 8x²

Therefore, either n + 1 or n - 1 must be a multiple of 4 so that n² - 1 is divisible by 8, implying that n should be an odd integer

So on trial, we have n = 3, 5, 7, .....

When n = 17:

17² = 8 x 6² + 1

So x = 6

y = 8x2 + 1

When x = 0, y
= 8(0) + 1
= 12
y is a perfect square

When x = 6, y
= 8(6)2 + 1
= 289
= 172
y is a perfect square

Ans: x = 0 or x = 6

2009-02-03 11:16:31 補充：
My answer should read as:

y = 8x^2 + 1

When x = 0, y
= 8(0) + 1
= 1
= 1^2
y is a perfect square

When x = 6, y
= 8(6)^2 + 1
= 289
= 17^2
y is a perfect square

Ans: x = 0 or x = 6