If a is an odd integer then show that a^2=1(mod 8)

a is odd then it can be written as
a = 2n + 1
so
a² becomes
a² = (2n + 1)² = 4n² + 4n + 1 = 4n(n + 1) + 1

n(n + 1) is even, because they are consecutive and one of the two factors is even
n(n + 1) = 2k

thus
a² = 4(2k) + 1 = 8k + 1
a²- 1 = 8k
that is
a² - 1 is a multiple of 8
a² ≡ 1 (mod 8)

QED

An odd integer is of the form 2n+1.
Its square will be (2n+1)² = 4n²+4n+1 = 4n(n+1)+1.
But either n or n+1 must be even. So n(n+1) is necessarily even
and 4n(n+1) is a multiple of 8. So...

Or you can reason as follows:
An odd number is necessarily congruent to 1, 3, 5 or 7 modulo 8.
The squares of those are 1, 9, 25 and 49
and they are all congruent to 1 modulo 8.

Let a = 2p + 1 be an odd integer.

Then a^2 = 4p^2 + 4p + 1 = 4p (p+1) + 1.

Since p is odd, p+1 is even, and we can call it 2q.

a^2 = 4p(2q) + 1 = 8pq + 1 = 1(mod 8)