# divisibility

• Sep 25th 2006, 12:12 AM
suedenation
divisibility
Hi, please take a look of the following question.

Show that if a^3 l b^2, then a l b.

Thanks for helping:)
• Sep 25th 2006, 12:36 AM
CaptainBlack
Quote:

Originally Posted by suedenation
Hi, please take a look of the following question.

Show that if a^3 l b^2, then a l b.

Thanks for helping:)

let p1^a1.p2^a2 ... pn^an be the prime decomposition of a, and
q1^b1.q2^b2. ... qm^bm be the prime decomposition of b.

That a^3|b^2, means that there is some qi which is equal to pj for
each of the pj's in the prime factorisation of a.

Also pj occurs with multiplicity 3aj in the prime decomposition of a^3, and
with multiplicity 2bi in the prime decomposition of b^2.

Hence we must have 2bi>=3aj, or (2/3)bi>=aj which means that bi>aj.

That is all the common primes in the prime decomposition of a and b
occur with greater multiplicity in b than in a which implies that a|b.

RonL
• Sep 25th 2006, 04:23 AM
ThePerfectHacker
Quote:

Originally Posted by suedenation
Hi, please take a look of the following question.

Show that if a^3 l b^2, then a l b.

Thanks for helping:)

Let,
gcd(a,b)=d
And let,
xd=a and yd=b
where,
gcd(x,y)=1
Thus,
a^3|b^2 becomes,
x^3d^3|y^2d^2
Thus,
x^3d|y^2
Thus,
y^2=kx^3d
Since,
gcd(x,y)=1
We cannot have that x has a non-trivial factor, i.e. besides for 1.
Thus,
x=1,
That means that,
gcd(a,b)=a
Thus,
a|b