# Prove that there are infinitely many primes of the form $6n + 5$

I know this is a duplicate but I had no idea what the other solution meant or how to go about approaching this problem. At first glance I thought about using Euclid's theorem that there are infinitely many primes and just doing proof by contradiction but I'm so confused right now. I'm horrible at math and I just want to understand how to approach this problem. I've tried proof by contradiction: Assume that there are only a finite number of primes in the form of $6n+5$. But I'm just completely lost on what to do after this.

The original question: How do you prove that there are infinitely many primes of the form $5 + 6n$?

• Please edit the question to include a link to the answer in the duplicate and explain to use just what you don't understand there. Without that information we would probably just suggest the same argument in an answer here. – Ethan Bolker Mar 1 '18 at 22:08
• My apologies, I just updated the question to include the original solution. – notGoodAtMath Mar 1 '18 at 22:11

## 3 Answers

First note that a prime of the form $6k+5$ is also of the form $6k-1$

Assume , there are only a finite many prime numbers of the form $6k-1$ and multiply them all together. The product must have residue $1$ or $5$ modulo $6$. If the residue is $1$, then subtract $2$. If the residue is $5$, then subtract $6$.

The resulting number is then greater than $1$ and of the form $6k+5$

It is easy to see that the resulting number cannot be divisible by $2$ or $3$, hence it must have prime factors of the form $6k\pm1$. If all prime factors were of the form $6k+1$, the number would be congruent to $1$ modulo $6$, which is not the case.

Hence, there must be a prime of the form $6k+5$ dividing the number, but because of the subtraction of $2$ or $6$ it cannot be one of the primes in the product. This gives you the desired contradiction.

• I'm afraid I'll be asking a stupid question, but why is a prime of the form 6k+5 also of the form 6k-1? Is this due to some fundamental theorems I'm missing or am I just overthinking things? – notGoodAtMath Mar 1 '18 at 22:22
• $6k+5=6k+6-1=6(k+1)-1$ – Peter Mar 1 '18 at 22:23

Note that the only primes not of the form $6n\pm 1$ are $2$ and $3$. A number of the form $6n+5$ is not divisible by $2$ or $3$.

Now note that the product $(6n+1)(6m+1)=36nm+6n+6m+1=6(6mn+m+n)+1$, and you can show by induction that any product of integers of the form $6n+1$ has the same form.

Any number of the form $6n+5=6(n+1)-1$ therefore has at least one prime factor of the form $6r+5$.

If you had a finite number of primes $p_i$ of the form $6r+5$ consider $n=4+\prod p_i^2$

You should be able to prove that this is of the form $6m+5$ and is not divisible by any of the $p_i$ (or by $2$ or $3$), but it is divisible by a prime of the form $6k+5$.

The essential step is showing that you can find a number of the form $6m+5$ which is not divisible by any of the $p_i$.

If we cross out from sequence of positive integers all numbers divisible by $$2$$ and all numbers divisible by $$3$$ then all remaining numbers will be in one of two forms:

$$S1(n)=6n+5=5,11,17,...$$ or $$S2(n)=6n+7=7,13,19,....n=0,1,2,3,...$$ So all prime numbers also will be in one of these two forms and ratio of number of primes in the sequence $$S1(n)$$ to number of primes in the sequence $$S2(n)$$ tends to be $$1$$. http://www.planet-source-code.com/vb/scripts/ShowCode.asp?txtCodeId=13752&lngWId=3