Question

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.

Answer 1

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

Answer 2

Mixing up my letters...:-)

Answer 3

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
\[m=p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdot ...\cdot p_n^{\alpha_n}\]

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

lol, that will do...

Answer 8

not the one you wanted i bet

Answer 9

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

Answer 10

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

Answer 11

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