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.

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, \(p\equiv 1\pmod{4}\).

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## Comments

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TopNewestFor 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.

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

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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, \(p\equiv 1\pmod{4}\).

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

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

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

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Do you mind explaining your solution a bit more?

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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.

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