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.
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Mathh Mathh
·
1 year, 1 month ago

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@Mathh Mathh
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This was the solution I was looking for. Good job!
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Daniel Liu
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1 year, 1 month ago

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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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Maggie Miller
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1 year, 1 month ago

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@Maggie Miller
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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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Daniel Liu
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1 year, 1 month ago

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@Daniel Liu
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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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Maggie Miller
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1 year, 1 month ago

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Note that number is (10^1008)^2+1. Since this is odd, the claim is true.
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Thinula De SIlva
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1 year, 1 month ago

@Daniel Liu
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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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Thinula De SIlva
·
1 year, 1 month ago

## 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. – Mathh Mathh · 1 year, 1 month ago

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– Daniel Liu · 1 year, 1 month ago

This was the solution I was looking for. Good job!Log in to reply

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 · 1 year, 1 month ago

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– Daniel Liu · 1 year, 1 month ago

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

– Maggie Miller · 1 year, 1 month ago

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

Note that number is (10^1008)^2+1. Since this is odd, the claim is true. – Thinula De SIlva · 1 year, 1 month ago

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– Daniel Liu · 1 year, 1 month ago

Do you mind explaining your solution a bit more?Log in to reply

– Thinula De SIlva · 1 year, 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.Log in to reply