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Mathematics
OpenStudy (anonymous):

Show that an integer (>1) with prime factors p1^k1p2^k2.....pn^kn is a square if and only if each of the exponents k is even.

OpenStudy (anonymous):

one way is easy enough. if \[k_i\] are even then this is the square of \[p_1^{\frac{k_1}{2}}\cdotp_2^{\frac{k_2}{2}}...\cdot p_n^{\frac{k_n}{2}}\]

OpenStudy (anonymous):

Mixing up my letters...:-)

OpenStudy (anonymous):

other way pretty straight forward too. if z is a square, then it is the square of something, say m by fundamental theorem of arithmetic \[m=p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdot ...\cdot p_n^{\alpha_n}\]

OpenStudy (anonymous):

And the generalization since u seem to be in fine form today...?

OpenStudy (anonymous):

so \[z=m^2=(p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdot ...\cdot p_n^{\alpha_n})^2\] etc

OpenStudy (anonymous):

one generalization is that \[\sqrt[n]{z}\] is irrational unless z is a perfect nth root

OpenStudy (anonymous):

lol, that will do...

OpenStudy (anonymous):

not the one you wanted i bet

OpenStudy (anonymous):

An integer is an nth power ifff each of prime exponents is a multiple of n (for lesse mortals)

OpenStudy (anonymous):

yeah so it is the same mutatis mutandis (god but i love fancy latin terms)

OpenStudy (anonymous):

U would make a number theorist cry....:-)

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