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. $\ddot\smile$

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Easy Math Editor

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Comments

A strange solution.

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

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

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

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.

Andrea Palma - 6 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 $4$ .

Haroun Meghaichi - 6 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$