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#3**0 **

for every proper integer factor of n that is not n^{1/2} there exists another proper integer factor of n such that their product is n

proof: if the proper factor is a, just divide n by a- n/a is a different proper factor, and a*(n/a)=n

this means that we can take the factors of n that are not the square root of n and pair two factors together if their product is n.

Suppose we have exactly K proper factors that are not the square root of n- this means that we have exactly K/2 (meaning that K is divisible by 2) pairs of proper factors. So when we'll multiply those factors of n (except for sqrt(n) of course) we'll get n^{K/2}.

why? because instead of multiplying all of the factors together, we can first find the product of each of the pairs, and then multiply the results together. but the product of each of the pairs is exactly n, and we have exactly K/2 pairs, therefore the answer is n^{K/2}.

*note- i did not assume that sqrt(n) is a factor of n- when n is not a perfect square, sqrt(n) is not a factor of n at all (in fact, it's irrational). But when sqrt(n) IS a factor of n, we cannot pair him like the other factors, because his "partner" will have to be n/sqrt(n)=sqrt(n), but because the members of a pair have to be different, i cannot allow that*

so, when n IS NOT a perfect square, sqrt(n) is not a factor of n, meaning that K=x (x=number of proper positive divisors)

, therefore the product of the proper positive divisors of n is n^{x/2}. We already know that the product is n^{(}^{ax+b)/c}, meaning that:

x/2=(ax+b)/c /multiply by c:

c*(x/2)=ax+b/ subtract ax from both sides:

(x/2)(c-2a)=b

Now, there is no single value for a+b+c- for example if x=6, a=-1, b=21 and c=5 we get x/2=(ax+b)/c and a+b+c=25

but a=1 b=0 and c=2 satisfies x/2=(ax+b)/c too, except this time a+b+c=3.

I'm not 100% sure i understood your question, are the numbers a b and c constants that satisfy n^{(ax+b)/c}=product of proper factors for every n, or just for a specific n? do a and b have to be positive?

i'm going to assume that you didn't mean that a b and c are constants, because it makes no sense.

but even when a, b, and c are have to be non-negative integers your question doesn't make sense.

i think i'll stop here.

please correct your question.

Guest Apr 25, 2018