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Home / Elementary Number Theory / Prove that 3a^2 – 1 is never perfect square.

Prove that 3a^2 – 1 is never perfect square.

Proof: Suppose 3a2 -1 = n2, some n.

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

So, n2 = 3k or, n2 = 3k + 1

⟹ 3a2 -1 = 3k or, 3a2 – 1 = 3k + 1

⟹ 3a2 – 3k = 1 or, 3a2 – 3k = 1+ 1

⟹ 3(a2 – k) = 1   or, 3(a2 – k) = 2

Each part is impossible.

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

So 3a2 – 1 is never perfect square.  (proved)

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