# OPC 2 Problem 1

$$1. \text{Prove that } 10^{2n + 1} - 1 \text{ is never a perfect square. For } n \in N$$

Give different methods of solving this. This was a problem to the OPC 2. The OPC 2 has already ended. OPC 3 will be released soon.Calvin sir suggested me to do this.So Thanks to him.$$\ddot\smile$$

Note by Rajdeep Dhingra
3 years, 2 months ago

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A strange solution.

Suppose $$10^{2n+1} - 1 = a^2$$

so

$$10^{2n+1} - 10 = a^2 - 9$$

$$10 \left(10^{2n} - 1\right) = (a - 3)(a+3)$$

LHS is even so RHS is even. But $$a-3$$ and $$a+3$$ are both even or both odd, so $$4$$ divides RHS.

But clearly $$4$$ doesn't divide LHS. Contradiction.

- 3 years, 2 months ago

A direct method would be : $$10^{2n+1}-1 \equiv -1 \pmod{4}$$. but a perfect square is $$0$$ or $$1$$ modulo $$4$$.

- 3 years, 2 months ago