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10 April, 01:08

Recall that the primes fall into three categories: Let Pi be the set of

primes congruent to 1 (mod 4) and P3 be the set of primes congruent to

3 (mod 4). We know that

{primes} = {2} UP, UP3.

We have previously proved that P3 is infinite. This problem completes

the story and proves that P1 is infinite. You can do this by following these

steps:

A) Fix n > 1 and define N = (n!) 2 + 1. Let p be the smallest prime divisor

of N. Show p>n.

B) If p is as in part (a), show that p ⌘ 1 (mod 4). (To get started, note

that (n!) 2 ⌘ 1 (mod p), raise both sides to the power p1 2 and go from

there. You will need Fermat's Theorem)

C) Produce an infinite increasing sequence of primes in P1, showing P1

is infinite.

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Answers (1)
  1. 10 April, 01:14
    0
    Check the explanation

    Step-by-step explanation:

    (a) Let p be the smallest prime divisor of (n!) ^2+1 if pn

    (b) (n!) ^2=-1 mod p now by format theorem (n!) ^ (p-1) = 1 mod p (as p doesn't divide (n!) ^2)

    Hence (-1) ^ (p-1) / 2 = 1 mod p hence [ as p-1/2 is an integer] and hence (p-1) / 2 is even number hence p is of the form 4k+1

    (C) now let p be the largest prime of the form 4k+1 consider x = (p!) ^2+1. Let q be the smallest prime dividing x. By the previous exercises q> p and q is also of the form 4k+1 hence contradiction. Hence P_1 is infinite
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