Prove $a, a + 2$, or $a + 4$ is composite?

Question

I've been assigned a proof by contrapositive for:

If $a$ is a natural number and $a, a + 2$, and $a + 4$ are prime, then $a = 3$.

So, the contrapositive I've produced is as follows:

If $a \neq 3$, then it is not true that $a, a + 2$, and $a + 4$ are prime.

How would I go about proving this?

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Mar 09 '20 at 08:50
Hint: Write $a= 3k + r$ where $r$ is the remainder of $a$ when divided by 3. — Leander Tilsted Kristensen, Mar 09 '20 at 08:51
This problem has been already posted in MSE. In https://math.stackexchange.com/questions/2055623/show-that-there-are-no-prime-triplet-other-than-3-5-7 where already was duplicate. — Piquito, Mar 09 '20 at 12:14

Qurultay · Accepted Answer · 2020-03-09T11:25:17.673

4

Let's do it looking at residue of $a$ modulo $3$:

If $a=3k$, for some $k>1$, then $a$ is not prime. Thus suppose $a\ne 3k$.

Now if $a=3k+1$, then $a+2=3k+1+2=3(k+1)$ is not prime.

If $a=3k+2$, then $a+2=3(k+1)+1$ and $a+4=3k+6=3(k+2)$ and $a+4$ is not prime.

Therefore, the only posibility for $a$ is $a=3$.

edited Mar 09 '20 at 11:25

answered Mar 09 '20 at 08:54

Qurultay

score 0 · Answer 2 · answered Mar 09 '20 at 11:30

0

Alternatively, you can use induction to prove that $a$, $a + 2$, or $a + 4$ is a multiple of $3$:

If $f(a)=a(a+2)(a+4)$, then $f(a+1)-f(a)=3 (a^2 + 5 a + 5)$.

answered Mar 09 '20 at 11:30

lhf

2 Answers2