- #1

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- Homework Statement
- By considering the number ## 4p_{1}^{2}p_{2}^{2}\dotsb p_{n}^{2}+3 ##, where ## p_{1}, p_{2}, ..., p_{n} ## are primes, prove that there are infinitely many primes of the form ## 6k+1 ##.

- Relevant Equations
- None.

Proof:

Suppose that the only prime numbers of the form ## 6k+1 ## are ## p_{1}, p_{2}, ..., p_{n} ##,

and let ## N=4p_{1}^{2}p_{2}^{2}\dotsb p_{n}^{2}+3 ##.

Since ## N ## is odd, ## N ## is divisible by some prime ## p ##,

so ## 4p_{1}^{2}\dotsb p_{n}^{2}\equiv -3\pmod {p} ##.

That is, ## (-3|p)=1 ##.

So ## p\equiv 1\pmod {6} ##, and hence ## p ## is one of the primes ## p_{i} ##.

But this is impossible, since ## p_{i}\nmid N ##.

This contradiction establishes the result.

Above is the proof for this problem in my book. But I do not understand why ## (-3|p)=1 ## for ## p\equiv 1\pmod {6} ##. Can anyone please explain why?

Suppose that the only prime numbers of the form ## 6k+1 ## are ## p_{1}, p_{2}, ..., p_{n} ##,

and let ## N=4p_{1}^{2}p_{2}^{2}\dotsb p_{n}^{2}+3 ##.

Since ## N ## is odd, ## N ## is divisible by some prime ## p ##,

so ## 4p_{1}^{2}\dotsb p_{n}^{2}\equiv -3\pmod {p} ##.

That is, ## (-3|p)=1 ##.

So ## p\equiv 1\pmod {6} ##, and hence ## p ## is one of the primes ## p_{i} ##.

But this is impossible, since ## p_{i}\nmid N ##.

This contradiction establishes the result.

Above is the proof for this problem in my book. But I do not understand why ## (-3|p)=1 ## for ## p\equiv 1\pmod {6} ##. Can anyone please explain why?