Assuming $a,k \in \mathbb{Z}$, if $(a-1)(a+1)=3k$ then why is it true that either $3|(a-1)$ or $3|(a+1)$?

Question

I was working on this problem:

Let $a$ be an integer. Prove $a$ is not evenly divisible by $3$ if and only if $a^2-1$ is evenly divisible by $3$.

I figured out how to prove the first statement, "if $3 \nmid a$ then $3|(a^2-1)$" by considering the cases where $a\equiv 1 \pmod{3}$ and $a\equiv 2 \pmod{3}$.

However, when I got to the converse, "if $3|(a^2-1)$ then $3\nmid a$" I got stuck, so I checked the solution manual for a hint, and it did the following to start out proving the converse:

Since $a^2-1$ is evenly divisible by $3$, then $a^2-1=3k,\exists k\in \mathbb{Z}$. Then $\color{red}{(a-1)(a+1)=3k\implies 3|(a-1)\: \text{or} \: 3|(a+1)}$.

After this, it proceeds to consider the cases where $3|(a-1)$ and $3|(a+1)$ to prove the statement. I just don't understand how we know the implication I've highlighted in red is true.

Also, now that I think about it, it may be more efficient to prove the converse using the contrapositive, "if $3|a$ then $3 \nmid (a^2-1)$" rather than this technique. — blakedylanmusic, Jan 31 '22 at 22:12
Because $3$ is prime, it divides a product $(a-1)(a+1)$ iff it divides one of the factors of that product. — Robert Shore, Jan 31 '22 at 22:19
Special case $,p=3,$ in Euclid's Lemma - see the linked dupe. — Bill Dubuque, Jan 31 '22 at 23:19

score 1 · Accepted Answer · answered Jan 31 '22 at 22:35

1

This is a direct result of Euclid's lemma:

If $a$ and $b$ are integers and $p$ is prime such that $p | ab$, then either $p | a$ or $p | b$ (or both).

answered Jan 31 '22 at 22:35

ConMan

24,300

Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jan 31 '22 at 23:19
Fair enough, I didn't see the dupe tag when I answered but I will keep an eye out in future. – ConMan Feb 01 '22 at 00:33

Assuming $a,k \in \mathbb{Z}$, if $(a-1)(a+1)=3k$ then why is it true that either $3|(a-1)$ or $3|(a+1)$?

1 Answers1