Fundamental Thm of Arithemetic

[1+1=1]

Wow, it has been awhile! China has been fun, but now it is time to get back to the states and also math work! Here is a problem that has been giving me some problems. It reads: [tex] \prod [/tex] from i=1 to n, pi^ai for each i is the canonical representation of a, deduce a formula for the sum of squares of the positive divisors of a. I know what the canonical representation is, so would i just plug in numbers for n, and then from the output just make up a [tex] \sum [/tex] formula? Could anyone provide guidance?

 

[shmoe]

Hi, are you familiar with the formula for the sum of the divisors of a? (not the squares). If so, this is a very closely related problem.

If not, here are the basic steps. Let f(a)=sum of squares of divisors.

1)find a formula when a is a prime power, that is if a=p^b, where p is prime, what is f(p^b)? If you need another hint, what are the divisors of p^b?

2)if m and n are relatively prime, show f(nm)=f(n)f(m), that is, f is multiplicative. If you have trouble with the general case here, try a simplified form first, where m and n are prime powers. This should help you see how to get the divisors of nm from the divisors of m and the divisors of n.

3)Combine the above to get a feneral formula for f(a) in terms of it's prime factorization.

 

[M98Ranger]

The Fundamental Theorem of Arithmetic is definitely an important concept to keep in mind when dealing with divisibility and prime factorization. It sounds like you have a challenging problem on your hands, but don't worry, I'm sure you'll figure it out!

To solve this problem, it may be helpful to start by breaking down the canonical representation of a into its prime factors. For example, if a = 12, then its canonical representation would be 2^2 * 3^1. From there, you can use the fact that the sum of squares of the divisors of a number can be calculated by taking the product of each prime factor raised to the power of 2, plus 1, and then multiplying all of those values together.

In the case of a = 12, the sum of squares of its divisors would be (2^2 + 1) * (3^1 + 1) = 3 * 4 = 12.

Hope this helps and good luck with your math work and transition back to the states!