Prove of disprove: If n is an integer with $n^2 = 3k$ for some integer k, then $n = 3j$ for some integer j.

Question

This was a problem on one of my recent discrete math exams and no one in the class was able to find the solution. My proof (which is definitely incorrect) is below. Any help would be appreciated.

Fact: Prove or disprove: If n is an integer with $n^2 = 3k$ for some integer k, then $n = 3j$ for some integer j.

Proof: If n is an integer with n^2 = 3k for integer k then n must be a factor of 3. Otherwise, its square would not yield 3k. This means that 3k = 9j. (3j)(3j) = 3k

Case: [n=6]

$n=6$

$n^2=36$

$k=12$

$j=2$

$36=3(12)$

$6=3(2)$

$3(12)=9(2)^2$

Is proven correct in the case of $n=6$.

Answer 1

Note that $$n\equiv 0,1,2\mod 3$$ so $$n^2\equiv 0,1 \mod 3$$

Answer 2

If $n^2=3k$, for some integer $k$, then $3\mid n^2$. Therefore, by Euclid's lemma, and since $3$ is prime, $3\mid n$.

Answer 3

By division algorithm, $n=3q+r$, where $r \in \{0,1,2\}$. Thus $$n^2=9q^2+6qr+r^2=3(3q^2+2qr)+r^2.$$ Since $3 \mid n^2$, this implies $3 \mid r^2$. But $r^2 \in \{0,1,4\}$ and the only number satisfying this property is when $r^2=0$, same as saying $r=0$. Thus $n=3q$ for some $q \in \Bbb{Z}$.