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

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.\(\ddot\smile\)

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TopNewestA 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.

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

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