# Fundamental Theorem of Arithmetic Help

• Feb 23rd 2007, 04:56 PM
jmc1979
Fundamental Theorem of Arithmetic Help
If anyone could help, I will appreciate it!

This are the problem:

1. Show that all of the powers in the prime-power factorization of an integer n are even if and only if n is a perfect square.

2. Show that every positive integer can be written as the product of possibly a square and a square-free integer. A "square-free integer" is an integer that is not divisible by any perfect squares other than 1.

Thank you!
• Feb 24th 2007, 06:11 AM
CaptainBlack
Quote:

Originally Posted by jmc1979
If anyone could help, I will appreciate it!

This are the problem:

1. Show that all of the powers in the prime-power factorization of an integer n are even if and only if n is a perfect square.

If N is a perfect square the there exists an n such that n^2=N.

Let n=p1^a1 p2^a2...pk^ak

be the prime decomposition of n, then:

N=n^2= p1^(2*a1) p2^(2*a2)...pk^(2*ak),

so the prime decomposition of N contains only even powers.

The other half of the proof goes through in a similar way, suppose
that the prime decomposition of N contains only even powers of primes,
then:

N = p1^(2*a1) p2^(2*a2)...pk^(2*ak),

so if n = p1^a1 p2^a2...pk^ak, N=n^2, so N is a square.

RonL
• Feb 24th 2007, 06:18 AM
CaptainBlack
Quote:

Originally Posted by jmc1979
2. Show that every positive integer can be written as the product of possibly a square and a square-free integer. A "square-free integer" is an integer that is not divisible by any perfect squares other than

Let the prime decomposition of an integer N be:

N = p1^a1 p2^a2...pk^ak = p1^[2*floor(a1/2)+(a1 mod 2)] p2^[2*floor(a2/2)+(a2 mod 2)] ... pk^[2*floor(ak/2)+(ak mod 2)]

.......= {p1^[2*floor(a1/2)] p2^[2*floor(a2/2)] ... pk^[2*floor(ak/2)]} {p1^[a1 mod 2] p2^[a2 mod 2] ... pk^[ak mod 2]}

.......= A B

where:

A = p1^[2*floor(a1/2)] p2^[2*floor(a2/2)] ... pk^[2*floor(ak/2)]

is a square, and:

B = {p1^[a1 mod 2] p2^[a2 mod 2] ... pk^[ak mod 2]}

is square free.

RonL
• Feb 24th 2007, 08:43 AM
jmc1979
Thank you!
Thank you for your help Capt. Black.
• Feb 24th 2007, 08:58 AM
jmc1979
Another one
This another problem:

Shot that if p^a||m then p^ka||m^k for any positive integer k.

I did this problem but dont know if the answer is correct.

Solution:
Suppose that p^a||m then we know that m=p^a X, where X is a prime other than p. Then if (m)^k=(p^a X)^k = m^k=p^ka X^k. Since p does not divide X^k for any positive integer k, then we know that p^ka||m^k.

Is this correct? Thank you to anyone!
• Feb 24th 2007, 03:05 PM
ThePerfectHacker
Quote:

Originally Posted by jmc1979
This another problem:

Shot that if p^a||m then p^ka||m^k for any positive integer k.

Just use the fact that if,
a|b and c|d then ac|bd.

Thus, for any positive k, we have k relations:

p^a | m
p^a | m
....
p^a | m

Multiply them out,
p^(ka)| m^k

Ah! I believe you are trying to show the converse, that is,
if p^k | m^k then, p | m.
(Which is not true always, but for primes, yes).
• Feb 24th 2007, 03:41 PM
jmc1979
Thanks
Thank you for the help!