How do I prove that every powerful number can be written as the product of a perfect square and a perfect cube?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

Think about it, what do you know about a square?

- anonymous

they are even

- anonymous

there you go thats all it is

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- KingGeorge

Just curious, but what class is this for?

- asnaseer

how does that prove it? I didn't know what "powerful numbers" were until I just looked them up.
if I understand it correctly then a powerful number is a positive integer m such that for every prime number p dividing m, p^2 also divides m.

- anonymous

Proofs class

- anonymous

I didn't really understand the question

- asnaseer

:/

- KingGeorge

I was curious because I helped out on the same question yesterday. See http://openstudy.com/study#/updates/50552fcde4b02986d370aedd

- anonymous

The first part makes sense but why are u subtracting 3 form ei when ei is odd?

- KingGeorge

So that I get \[\large p_i^{e_i}=p_i^{f_i+3}=p_i^{f_i}\cdot p_i^3\]Note that since \(e_i\) is odd, \(e_i-3\) is even, so \(\displaystyle p_i^{e_i-3}=p_i^{f_i}\) is a perfect square.

- KingGeorge

Additionally, \(p_i^3\) is a perfect cube.

- anonymous

oh that makes perfect sense, thanks alot

- KingGeorge

You're welcome.

Looking for something else?

Not the answer you are looking for? Search for more explanations.