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Math Help - divisibility proof

  1. #1
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    divisibility proof

    In this thread:
    http://www.mathhelpforum.com/math-he...an-triple.html

    I was trying to prove that if d^2 divides z^2, then d divides z.

    It seems like it should be an easy proof, but the proof I came up with is long and unclear (maybe even wrong). Can anyone come up with a better proof?

    Here it is:
    By definition, d^2|z^2 means there is a positive integer n such that z^2=nd^2. Suppose n is not the square of an integer. Let n=ms^2, where m>1 has no square divisors. Let p be a prime divisor of m. We have z^2 = nd^2 = ms^2d^2, and since the powers of p in z^2, s^2, and d^2 are even, the power of p in m must be even. But then m has a square factor. By contradiction, therefore, n must be the square of an integer. Let n=a^2. Then z^2=a^2d^2 and since z and d are positive and you can choose a to be positive, z=ad, which is the definition of d|z.
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  2. #2
    Senior Member Tinyboss's Avatar
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    It'll be easier if you think in terms of prime factorizations, i.e. if

    d=p_1^{s_1}p_2^{s_2}\cdots p_n^{s_n} and z=q_1^{r_1}q_2^{r_2}\cdots q_m^{r_m}, with all p's and q's prime, and all s's, r's positive integers,

    how would you express the relationships d\mid z and d^2\mid z^2?
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  3. #3
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    So:
    d|z means that if p_i=q_j then s_i\le{r_j}
    and:
    d^2|z^2 means that if p_i=q_j then 2s_i\le{2r_j}
    which are obviously equivalent statements.

    I was looking for something more like the proof of a|b\text{ and }a|c \rightarrow a|(mb+nc) for any integers m,n. From the definition of divisibility, there are integers r,s such that b=ra and c=sa. Then mb+nc = (mr+ns)a and so a divides mb+nc.

    Maybe what I'm looking for doesn't exist...
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