# Proof: Suppose 3a^{2} -1 = n^{2}, some n.

We know the square of any integer is either of the form 3k or 3k + 1.

So, n^{2} = 3k or, n^{2} = 3k + 1

⟹ 3a^{2} -1 = 3k or, 3a^{2} – 1 = 3k + 1

⟹ 3a^{2} – 3k = 1 or, 3a^{2} – 3k = 1+ 1

⟹ 3(a^{2} – k) = 1 or, 3(a^{2} – k) = 2

Each part is impossible.

Since by division algorithm, 2 = 3.0 + 2 and 1 = 3.0 + 1

*So 3a ^{2} – 1 is never perfect square. (proved)*