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In an exercise I'm asked to prove the following:

Let $3k + 2$ be a natural number. Prove that, $3k + 2$ has a prime factor of the same form.

So I was able to prove that, if $k$ is an even number, then $2=3\cdot0+2\mid 3k+2$ but I'm not being able to conclude anything if $k$ is an odd number. How can I prove that, if $k$ is an odd number then exists $k'$ such that $3k'+2$ is prime and $3k'+2\mid 3k+2$?

Bernard
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Eduardo Magalhães
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    Try a proof by contradiction: what would the form of all the factors of this number be if the statement is false? – player3236 Jan 02 '21 at 20:33
  • More generally see the proof of this Lemma - which is used in generalizing Euclid's proof that there are infinitely many primes (put $,S = 3,\Bbb N + 1,\ c(n) = 3n-1)\ \ $ – Bill Dubuque Jan 02 '21 at 21:41

3 Answers3

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Note that a prime $p$ dividing $3k+2$ cannot be $0\pmod{3}$. Then, every prime dividing $3k+2$ is either $1\pmod{3}$ or $2\pmod{3}$.

If all primes dividing $3k+2$ were $1 \pmod{3}$ then their product (regardless of their power) would be $\equiv 1 \pmod{3}$ which gives a contradiction.

Aven Desta
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Sofía Marlasca Aparicio
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Hint:

All primes $\ne 3$ are congruent to $1$ or $-1\mod 3$, and $n\equiv -1 \mod 2$.

Bernard
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PROOF BY CONTRADICTION:

Hint: the product of any two integers of the form $(3m+1)$, is $(3m+1)$

Solution: if $p_1$, $p_2$, ... are prime factors of the number, and if $p_1 \equiv 1 \pmod{3}$, $p_2 \equiv 1 \pmod{3}$ ..., then $p_1 \cdot p_2 \cdot ... \equiv 1 \pmod{3}$

Aven Desta
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