Question

Show that if a and b are positive integers and a^3 | b^2, then a| b
Please, help

Answer 1

have you thought about using fundamental theorem of arithemetic maybe?

Answer 2

I don't see the result from it. I tried to use
\(Let ~ a =\prod p_i^{a_i}\)

\(a^3 = \prod p_i^{3a_i}\)
Same as b
\(b^2=\prod p_i^{2b_i}\)
(2,3)= 1, hence if \(a^3 | b^2\) only one case can happen that is \(a_i | b_i\), hence a|b
But I don't feel positive on this solution.

Answer 3

I think you will need to use induction. You would be able to prove this, but the question says show...

Answer 4

What you have
a^3 | b^2 which means k*a^3 = b^2 where k is an integer


What you want to show
a | b, in other words m*a = b where m is an integer


Hint:
k*a^3 = b^2
k*a*a^2 = b^2
(k*a)*(a^2) = b^2
q*(a^2) = b^2 ... let q = k*a

Answer 5

@jim_thompson5910 does k need to be positive also?

Answer 6

You meant: \(ka^3 = b^2 \\(ka)a^2 = b*b\)
either ka | b or a^2 | b
If ka |b , we are done
if a^2 | b, then?? \(b = ma^2 = ma* a \) and we done , right?

Answer 7

a,b are positive
b^2, a^3 are positive
if k is negative, then k*a^3 is negative which contradicts the fact that b^2 is positive

so yes, k has to be positive

Answer 8

okay

Answer 9

Still not get @jim_thompson5910 please, guide me more

Answer 10

The RHS is a perfect square and an integer
so the LHS must also be a perfect square and an integer

From these steps
q*(a^2) = b^2
q = (b^2)/(a^2)
q = (b/a)^2
we can see q is a perfect square

Answer 11

Can I do:
a^3 | b
a^3 | b*b
then a^3 | b
b = ka^3 = aka^2
a|b
is it right?

Answer 12

if q is a perfect square, then q = m^2 for some integer m

q*(a^2) = b^2 ... let q = k*a
(m^2)*(a^2) = b^2 ... let m^2 = q

Answer 13

Got you. Thanks a lot.

Answer 14

no problem