Prime factors of 10...01

Prove that all prime factors $p$ of $1\underbrace{00\cdots 00}_{2015\text{ 0's}}1$ are of the form $p\equiv 1\pmod{4}$ .

#NumberTheory #ModularArithmetic #PrimeFactorization #PrimeFactors #LegendreSymbol

Easy Math Editor

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`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}`	$\boxed{123}$

Comments

For contradiction, let $p\equiv 3\pmod{4}$ satisfy $\left(10^{1008}\right)^2\equiv -1\pmod{p}$ .

Raise both sides by $(p-1)/2$ (which is odd); we get $\left(10^{1008}\right)^{p-1}\equiv (-1)^{(p-1)/2}\equiv -1\pmod{p}$ , which contradicts Fermat's Little Theorem.

mathh mathh - 5 years, 9 months ago

This was the solution I was looking for. Good job!

Daniel Liu - 5 years, 9 months ago

Note $n=100\cdots001=10^{2016}+1=(10^{1013})^2+1^2$ . By Fermat's theorem on the sum of two squares, any prime factor $p$ of $n$ satisfies $p\equiv1\pmod{4}$ or $p=2$ . Since $n$ is odd, $p\equiv 1\pmod{4}$ .

Maggie Miller - 5 years, 9 months ago

Okay, I was hoping for a solution that didn't use Fermat's two square theorem. Thanks regardless.

Daniel Liu - 5 years, 9 months ago

I thought it was kind of clever! Haha, sorry that wasn't what you were looking for.

Maggie Miller - 5 years, 9 months ago

Note that number is (10^1008)^2+1. Since this is odd, the claim is true.

Thinula De SIlva - 5 years, 9 months ago

Do you mind explaining your solution a bit more?

Daniel Liu - 5 years, 9 months ago

Well Fermat said that if n is even, all prime factors of n^2+1 are 1mod4. Thus, we can see that 10^1008 is obviously even, so we're done.

Thinula De SIlva - 5 years, 9 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}`	$\boxed{123}$