Question

Show that for odd positive integer to be a perfect square, it should be of the form 8k + 1

Answer 1

Any odd positive integer is of the form
4x + 1 or 4x + 3
where 'x' is any integer.
Let y = 4x + 1
Squaring on both sides, we get:
y² = (4x + 1)²
y² = 16x² + 8x + 1
y² = 8 x (2x + 1) + 1
y² = 8 k + 1
where = k = x (2x + 1)
Now let,
y = 4x + 3
Squaring on both sides, we get:
y² =(4x + 3)²
y² = 16x² + 24x + 9
y² = 16x² + 24x + 8 + 1
y² = 8 (2x² + 3x + 1) + 1
y² = 8 k + 1
where k = 2x² + 3x + 1
which shows that the square of any positive odd integer is of the form 8 k + 1 where k is any integer.
Hopefully it helps.
Thanks.