Prove that if $5|n^3$ then $5|n$. I have this very basic number theory type proof that is causing me some headaches and I know it is probably very simple.
I want to prove that if $5|n^3$ then $5|n$.
I have a condition that I am imposing here, one cannot use GCD function.
It is quite obvious to me that $5|n$, due to the fact that $n^3$ is $n*n*n$, and when dividing by $5$, its obvious that it must divide into one of the duplicate '$n$' here.
I am not sure if this can be done via a contrapositive proof or proof by contradiction. It seems to me that it can be done via a direct proof.
So if someone can help with a proof of this, would really appreciate it.
 A: Yes, From Euclid's Lemma if $p$ is a prime number and $p$ divides $a.b$ , $a,b$ positive integers then $p|a$ or $p|b$.
Here $p = 5 | n^3$ or $p | n$ or $p|n^2$.
If $p|n$ we are done!
For the other case of $p|n^2$ then $p|n$ or $p|n$ so we are done again :)!
A: Based on the level of your question, I suppose you don't know a lot of facts about prime numbers. So this one is most easily handled by contrapositive. Suppose $5 \nmid n$, meaning that $n = 5k + j$ for some integers $k$ and $j$, with $j \in \{1, 2, 3, 4\}$ (why can I assume this??). 
Then work through four cases, computing $n^3$. For example,
$$(5k + 1)^3 = 125 k^3 + 75k^2 + 15k + 1 = 5(25k^3 + 15k^2 + 3k) + 1$$
is not divisible by $5$.
A: $5$ is a prime number and $n^3=n\cdot n\cdot n$.
Thus, $n$ is divided by $5$.
If $n$ is not divided by $5$ then $n=p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k},$ where $\alpha_i\in\mathbb N\cup\{0\}$,  $p_i$ are primes and $p_i\neq5$.
Thus, $n^3=p_1^{3\alpha_1}p_2^{3\alpha_2}...p_k^{3\alpha_k}$ is not divisible by $5$, which is a contradiction.
A: if $5|n^3$ and 5 doesn't divide n then we must have $5|n^2$
$$\frac {n³}{5}= n\left(\frac {n^2}{5}\right)$$
if $5|n^2$ and 5 doesn't divide n then we must have $5|n$
$$\frac {n^2}{5}= n\left(\frac {n}{5}\right)$$
Contradiction...
So if $5|n^3$ then $5|n$
