$3\mid n,m\Rightarrow 3\mid n^2, jn+km\,$ Multiples closed under scalings and addition.

Question

Suppose $n$ is an integer. If 3|n then 3|$n^2$. Prove.

So I'm wondering if both approaches here are ok.

$1^{st}$:

3|n so $n=3a$ a in integers

$n^2=9a^2$

$n^2=3(3a^2)$

$2^{nd}$:

$n=3a$

(both cases we have 3 times an integer) Thanks

$n^2=3an$

Answer 1

Both are correct. The 2nd is essentially $\ 3\mid n\mid nm\,\Rightarrow\,3\mid nm\ $ by transitivity of "divides" (OP is case $\,m = n)$. Thus the set $\,3\Bbb Z\,$ of multiples of $3$ is closed under multiplication by any integer $m$. Similarly you can show it is closed under addition (so also subtraction by $\,a-b = a+(-1)b\,]$. Hence we infer that $\, \bbox[5px,border:1px solid #c00]{3\mid m,n\,\Rightarrow\, 3\mid j m + k n}\,$ for all integers $\,j,k,\,$ with similar proof, i.e.

$$ m,n = 3a,3b\,\Rightarrow\, jm\!+\!kn = j(3a)\!+\!k(3b) = 3(ja+kb)\qquad$$

Clearly the proof remains valid if we replace $3$ by any integer. This linear structure of the set of multiples of an integer (or common multiples of many integers) lies at the heart of many divisibility results. This innate linear structure is brought to the fore when one studies ideals in rings, as well as modules - a generalization of a vector space with scalars from rings (vs. fields).