4

Argue that there are infinitely many primes p that are not congruent to 1 modulo 5.

I find this confusing. Is this saying $p_n \not\equiv 1 \pmod{5}$?

To start off I tried some examples.

$3 \not\equiv 1 \pmod{5}$

$5 \not\equiv 1 \pmod{5}$

$7 \not\equiv 1 \pmod{5}$

$11 \equiv 1 \pmod{5}$

$13 \not\equiv 1 \pmod{5}$

$17 \not\equiv 1 \pmod{5}$...

If this is what the question is asking I've come to the conclusion that this is true. Either way, I've got no clue how to write this as a proof.

Martin Sleziak
  • 53,687
maroon.elephants
  • 249

2 Answers2

10

You can follow the Euclid proof that there are an infinite number of primes. Assume there are a finite number of primes not congruent to $1 \pmod 5$. Multiply them all except $2$ together to get $N \equiv 0 \pmod 5$. Consider the factors of $N+2$, which is odd and $\equiv 2 \pmod 5$. It cannot be divisible by any prime on the list, as it has remainder $2$ when divided by them. If it is prime, we have exhibited a prime $\not \equiv 1 \pmod 5$ that is not on the list. If it is not prime, it must have a factor that is $\not \equiv 1 \pmod 5$ because the product of primes $\equiv 1 \pmod 5$ is still $\equiv 1 \pmod 5$

Martin Sleziak
  • 53,687
Ross Millikan
  • 374,822
  • I don't think you can get away easily by adding $2$ or $3$ because they ARE divisible by primes that are not $1 \bmod 5$. – fretty Nov 06 '12 at 18:52
  • @fretty I second that. I don't immediately see how to deal with 2 and 3. – Dan Shved Nov 06 '12 at 18:53
  • You would have to show that this number is not a perfect power of the number you added. – fretty Nov 06 '12 at 18:55
  • @fretty still don't see how this is enough. What if it is a product of the number I added and several other primes that are 1 mod 5? – Dan Shved Nov 06 '12 at 18:57
  • I think there is also a problem with this number being divisible by $5$. – fretty Nov 06 '12 at 18:57
  • @Dan Shved: That is not a problem, the contradiction comes from $p_i |2$ or $p_i|3$ for some $i$. You thus have to show that the constructed number is NOT JUST divisible by such a prime. – fretty Nov 06 '12 at 18:58
  • My advice is to prove something stronger using Ross' advice. Prove that there are infinitely many primes congruent to $4 \bmod 5$. – fretty Nov 06 '12 at 19:03
  • @fretty Could you expand on that (about powers of 2)? I took the product $N$ of all the primes who are not congruent to 1 (mod 5). Now I am looking at the number $N+2$, which is congruent to 2 (mod 5). Then I should look at the prime factors of $N+2$. And what do I do about them, how do I proceed? – Dan Shved Nov 06 '12 at 19:04
  • Well the way the usual proof goes is that you want there to exist a prime factor of N+2 that is not $1 \bmod 5$, then it must be one of the primes on your finite list and so divides $2$. The "and so it divides 2" bit is usually the contradiction but here your prime could actually be 2 for all you know! – fretty Nov 06 '12 at 19:06
  • Oh but I have been silly because we may assume N+2 to be odd! – fretty Nov 06 '12 at 19:08
  • It's actually even. – Dan Shved Nov 06 '12 at 19:08
  • 2
    Well we can discard the prime $2$ and consider only odd primes, it doesn't change the outcome of the theorem. – fretty Nov 06 '12 at 19:09
  • @fretty OK, that seems to work. – Dan Shved Nov 06 '12 at 19:12
  • @fretty: right you are. I have updated it. – Ross Millikan Nov 06 '12 at 19:12
  • $N+2$ might actually be $1$ ($\bmod$ $5$). – WimC Nov 06 '12 at 19:26
  • @WimC: no, because $N \equiv 0 \pmod 5$ as 5 is a prime. – Ross Millikan Nov 06 '12 at 19:27
  • @RossMillikan: fair enough. – WimC Nov 06 '12 at 19:30
  • @Ross I think you should add further details to the proof. I cannot tell which way you intend it to be completed. – Bill Dubuque Nov 06 '12 at 19:54
  • @BillDubuque: is this what you were thinking? – Ross Millikan Nov 06 '12 at 20:39
2

Hint $\rm\ \ 5n^2\!-n\: $ has a larger set of prime factors $\rm\not\equiv 1\ mod\ 5\:$ than does $\rm\:n.$

Bill Dubuque
  • 272,048