# If $p$ is prime and $\gcd(m,p) = 1$ show that $\gcd(m,p^k) = 1$

If $p$ is prime and $\gcd(m,p) = 1$ show that $\gcd(m,p^k) = 1$ where $k\geq1$.

I think I have come up with a solution:

Suppose $m$ has prime factorization $m = p_1^{a_1}...p_n^{a_n}$ where $p_i$ is prime. Since $\gcd(m,p) = 1$ none of $p_i = p$. Now if $\gcd(m,p^k) \neq 1$ then the $\gcd$ must be $p^n$ with $0< n \leq k$. Note that the divisors of $m$ are the integers $p_1^{\bar{a_1}}...p_n^{\bar{a_n}}$ with $0\leq \bar{a_i} \leq a_i$.

So then we must have that $p^n = p_1^{\bar{a_1}}...p_n^{\bar{a_n}}$ so then both numbers must have the same prime factorization so that means that one of $p_i = p$ a contradiction. Which proves the result.

Is this argument fine?

• Looks good. Longer than necessary. If $\gcd(m,p^k)\gt 1$, then some prime $q$ divides $m$ and $p^k$. Argue that $q=p$ from Euclid's Lemma, if a prime $q$ divides a product it divides one of the terms. Nov 14 '15 at 18:55

Since $$\gcd(m,p)=1$$, $$p$$ cannot divide $$m$$ (otherwise $$p$$ would be a greater common divisor).
The only divisors of $$p^k$$ are $$p^n$$ for $$0\le n\le k$$ and if any of those except $$1$$ divides $$m$$, then $$p$$ will also divide $$m$$, which we know it doesn't. So $$1$$ is the only (positive) common divisor of $$m$$ and $$p^k$$.