Prove:
If p is prime and $p|c^2$ then $p|c$ and $p^2|c^2$.

Question

Prove:
If p is prime and $p|c^2$ then $p|c$ and $p^2|c^2$.

thomas5267 · Answer

Can I argue that $c$ factors into unique primes $p_1^{a_1}p_2^{a_2}\ldots p_n^{a_n}$and $c^2=p_1^{2a_1}p_2^{2a_2}\ldots p_n^{2a_n}$ so if $p|c^2$ then $p|c$ as the exponent are all even and by extension $p^2|c^2$?

thomas5267 · Answer

Please check my prove.  @ganeshie8

ganeshie8 · Answer

that works but i think a proof without calling for prime factorization gives more insight into other properties...

ganeshie8 · Answer

Save prime factorization proof, I would try proof by contradiction as it tells me what I need to start with atleast..

ganeshie8 · Answer

Prove 
If $p|c^2$ then   $p|c$ and  $p^2|c^2$

Proof (by contradiction) :
Suppose $p|c^2$ but $p \nmid  c$, then we have  $c = pn+r$ for some $0\lt r \lt p$
Since $p|c^2$, we have $c^2 = pm$ for some integer $m$ 
$\implies  (pn+r)^2 = pm$
$\implies  p(pn^2 + 2nr)+r^2 = pm$
which is impossible unless $r=0$. contradiction. $\blacksquare$

ganeshie8 · Answer

Contradiction occurs because we have assumed $r$ is non zero which is required for $p \nmid  c$