Question

show that the odd positive integer to be perfect square it should be of the form 8-k + 1​

Answer 1

sorry how can it come like that question is wrong

Answer 2

Any odd integer can be of the form (4a + 1) and (4a + 3). (where a is an integer)

The square of every odd integer is an odd integer.

Therefore, every perfect square odd integer can be of two forms:

1. (4a + 1)²                              or                                   2. (4a + 3)²

Case 1:

(4a + 1)² = 16a² + 8a + 1.

             = 8(2a² + a) + 1

If k = 2a² + a;

(4a + 1)² = 8k + 1

Case 2:

(4a + 3)² = 16a² + 24a + 9

              = 16a² + 24a + 8 + 1

              = 8(2a² + 3a + 1) + 1

If k = 2a² + 3a + 1;

(4a + 3)² = 8k + 1

Hence proved.

Hope this helps!