×

# 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
2 years, 9 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

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.

- 2 years, 9 months ago

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

- 2 years, 9 months ago