Question

For an odd positive integer to be perfect square it should be necessarily be of the 8k + 1 unable to understand this line

Answer 1

Here is the answer to your question.
Any odd positive integer ( eg : 3 = (4×0+3) , 5 = (4×1+1)) 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)

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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.