Question

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

Answer 1

Here is the answer to your question.
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.

Answer 2

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