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Math Help - Proof... please help.

  1. #1
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    Proof... please help.


    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?
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  2. #2
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    Quote Originally Posted by Dreamer78692 View Post

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