Can you prove that if $3 \mid 2a$, then $3 \mid a$, with prime factorization?

Question

The statement goes as following: if $3 \mid 2a$, then $3 \mid a$ and $a$ is an integer. In my approach, I used prime factorization, but is this actually valid? This was my approach:

$$3 \mid 2a \implies 2a = 2 \cdot 3 \cdot k, k \in \mathbb{N}$$

$$\frac{2a}{2} = \frac{2\cdot 3 \cdot k}{2}$$

$$a = 3 \cdot k$$

$$\therefore 3 \mid a$$

Is this valid or am I doing forbidden things here?

Answer 1

Note that $2a=3k$ implies $$ a=3a-2a=3(a-k)$$

Answer 2

First note that 1 = 3 - 2. Multiply by a: a = 3a - 2a. If 2a = 3n (for some n), then a = 3a - 3n = 3(a - n). Hence 3 | a.

This generalizes to other prime pairs: When p, q are distinct primes, GCD(p,q) = 1, hence there are (signed) integers m, n such that 1 = pm + qn. Multiplying both sides by 'the other value in the product' (here a) isolates that value on the left and produces an expression divisible by p or q (depending on which is 'the first value in the product') on the right.

This also generalizes to algebraic structures more general than numbers. It does NOT work when the dividing item (here 3) is NOT prime!

Answer 3

Yes: $\,\ 3\mid 2a,3a\,\Rightarrow\, 3\mid 3a\!-\!2a = a.\ $ QED $\ $ We used Bezout $3 - 2 = 1.\,$ This generalizes:

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Generally $\,c\mid ab\,\Rightarrow\, c\mid a\ $ if $\ b,c\,$ are (Bezout) coprime $\ jb+kc = 1\,$

Proof $\ \ c\mid abj,akc\,\Rightarrow\, c\mid a(jb+kc) = a\ \ $ QED

More generally, using basic gcd laws (notably gcd distributive law)

$\qquad\quad\ c\mid ab,ac\ \Rightarrow\ c\mid (ab,ac) = a(b,c) = a$

Note that none of the above proofs use prime factorization.