The number of integers n with 1 ≤ n ≤ 100 for which n^2 ≡ 1 ±od8 is

The correct answer is Option (3) → 50

Let us look at even and odd values of \(n\) separately.

For an odd number \(n = 2k + 1\):

    \(n^{2} = 4k^{2} + 4k + 1 = 4k(k + 1) + 1\)

One of \(k\) and \(k + 1\) is even, so \(k(k + 1)\) is even and \(4k(k + 1)\) is a multiple of 8.

    \(n^{2} \equiv 1 \pmod{8}\) for every odd \(n\)

For an even number \(n = 2k\):

    \(n^{2} = 4k^{2} \equiv 0 \text{ or } 4 \pmod{8}\)

So exactly the odd numbers work. From 1 to 100 there are 50 odd numbers.