Prove

**footmath** · January 7th 2012, 07:18 AM

Prove that there is no perfect square such that the left number is

and the right number is

and other numbers is

.

**Bacterius** · January 7th 2012, 09:26 AM

Let's see... 301... 3001...30001, can we describe this class of integers with a formula? Yes!

Basically you have to prove that $3 \times 10^n + 1$ is not a perfect square for all $n > 1$ (or $n > 0$ , whatever).

Thus $x^2 = 3 \times 10^n + 1$ has no solution for $x \in \mathbb{N}$ for all $n > 1$ .

The above equation is equivalent to $x^2 - 1 = 3 \times 10^n$

... also known as $(x - 1)(x + 1) = 3 \times 10^n$

Can you take it from there?

**footmath** · January 12th 2012, 09:57 PM

Originally Posted by Bacterius

Let's see... 301... 3001...30001, can we describe this class of integers with a formula? Yes!

Basically you have to prove that $3 \times 10^n + 1$ is not a perfect square for all $n > 1$ (or $n > 0$ , whatever).

Thus $x^2 = 3 \times 10^n + 1$ has no solution for $x \in \mathbb{N}$ for all $n > 1$ .

The above equation is equivalent to $x^2 - 1 = 3 \times 10^n$

... also known as $(x - 1)(x + 1) = 3 \times 10^n$

Can you take it from there?

Thanks I am not well in number theory.
How can we prove that $3 \times 10^n + 1$ is not a perfect square for all [TEX]n > 1 ?
In other method can we write 301... 3001...30001 as $8t+k$ and say because it cannot be written as $8q+1$ is not perfect square ?

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