If pq|(n^2) proof that pq|n...
p is prime and q is prime and pis not equal to q.

My proof I used the contrapositive to proof it..
if pq not divisble by n then pq is not divisible by n squared
*!= is not equal to..
Short version of proof
n!=pqk where k is element of Z
(n)^2 !=pqc where c=kpq is element of Z

please check if this is right...
How will you prove this directly?

2. Originally Posted by Dreamer78692

If pq|(n^2) proof that pq|n...
p is prime and q is prime and pis not equal to q.

My proof I used the contrapositive to proof it..
if pq not divisble by n then pq is not divisible by n squared
*!= is not equal to..
Short version of proof
n!=pqk where k is element of Z
(n)^2 !=pqc where c=kpq is element of Z

please check if this is right...
How will you prove this directly?

Please note that $pq\mid n^2$ means " $pq\,\, divides\,\,n^2$ or $n^2\,\,is\,\,divisible \,\,by\,\,pq$ , and not the other way around.

Now, you only wrote the contrapositive of your problem, but you didn't provide any proof...

I'd go like this: write n as a producto of powers of primes: $n=r_1^{a_1}\cdot\ldots\cdot r_s^{a_s}\,,\,\,r_i\,\,primes\,,\,\,a_i\in\mathbb{ N}$ ,so:

$pq\mid n^2\Longrightarrow n^2=r_1^{2a_1}\cdot\ldots\cdot r_s^{2a_s}=kpq\,,\,\,k\in\mathbb{Z}$ , and from the uniqueness (up to order) of

the representation as a product of primes, it must be that $p\mid r_i\,,\,q\mid r_j$ , for some indexes $i,j$ . Take it from here.

Tonio

3. I cant believe I never think of that thanks man....