Use $(a,b)=1$ and $a | bc$ then $a | c$ to show that if $p$ is prime and $a$ is an integer with $p | a^2$, then $p | a$

Question

problem : Use $(a,b)=1$ and $a | bc$ then $a | c$ to show that if $p$ is prime and $a$ is an integer with $p | a^2$, then $p | a$

What I've done: Use contradiction that assume $p$ does not divide $a$, then $(p, a) = 1$

and I am stuck here....

what is the next step or how can i prove this question?

Then $(p,a) = 1$, but $p \mid a^2 = a \times a$ , so $p | a$ (in the lemma, take $b=c=a$),contradiction. — Sarvesh Ravichandran Iyer, Sep 24 '17 at 05:12
Let $p=A^$ and $a=B^=C^. $ Then $A^|B^C^$ and $A^$ is prime so if $A^\not |; B^*$ then .....? — DanielWainfleet, Sep 24 '17 at 05:26

score 0 · Accepted Answer · answered Sep 24 '17 at 05:53

Do what it says:

Prove the following lemma:

Lemma: If $\gcd(a,b) = 1$ and $a|bc$ then $a|c$.

Proof: .... that's for you to do....

Then use that to prove if $p$ is prime and $p|a^2$ then $p|a$>

Either $\gcd(a,p) = 1$ or $\gcd(a,p) = p$. If $p|a^2=a*a$ and $\gcd(a,p) = 1$ then $p|a$ which contradicts $\gcd(a,p) = 1$. So $\gcd(a,p) = p$ and $p|a$.

That's it.

Okay... proof of the lemma.... If $a|bc$ then $\frac {bc}a$ is an integer. So all the factors of $a$ divide into $bc$. Bt $\gcd(a,b) = 1$ so $a$ and $b$ have no factors in common. So none of the factors of $a$ divide into $bc$. So all the factors of $a$ divide into $c$. So $a$ divides $c$.

Use $(a,b)=1$ and $a | bc$ then $a | c$ to show that if $p$ is prime and $a$ is an integer with $p | a^2$, then $p | a$

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