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# 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 $$p\equiv 1\pmod{4}$$.

Note by Daniel Liu
2 years, 1 month ago

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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.

- 2 years, 1 month ago

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

- 2 years, 1 month 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, $$p\equiv 1\pmod{4}$$.

- 2 years, 1 month ago

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

- 2 years, 1 month ago

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

- 2 years, 1 month ago

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

- 2 years, 1 month ago

Do you mind explaining your solution a bit more?

- 2 years, 1 month 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.

- 2 years, 1 month ago