Lets say I have the value of c...

Lets say I have the value of c. Is there a way to find all possible pairs of natural numbers that could be put under a and b to give c?

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No

By brute forcing it, sure.

you would find all the factors of C

every factor of C is an A value

then simply plug in and solve

EZ PZ hope u get a good grade on your homework faggot

We can easily determine that a lies inbetween (2,c) and b lies inbetween (1,logbase2 c). I don't know how to go any further from this.

was about to post what this user said. thats the gist of it. There might be a better way to do it though and obviously only works for natural number solutions.

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this is basically how simple cryptography works (cryptographic hashing functions use prime numbers though). operations in one direction are fast (calculating c). operations in the other direction (find a and b given c) are slow.

myt8n

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and do add, the factors are all CANDIDATES for the a value. Some won't work, for example. for c=6, the factors are 3, 2, and 6. Only 6 works here.

im curious does anyone have a better solution than factoring and checking the factors?

Is c a natural number?
Do you count 0 as one of the natural numbers?
If so, is c ever 0?

A, B, and C area all natural numbers.

and 0 is not a natural number.

I think this is an outstanding open problem, isn't it?

Nvm, I miss read. I thought op was asking something else. Idk why.

what did you read it as?

well you don't have to check any factor, prime or otherwise, greater than the bth root so you can save a little time there

I thought op was asking how large c could be given a and b as distinct natural numbers. Idk why.

Im pretty sure there there would be no limit there.

The set of natural numbers is infinite. Doesn't that mean that a and b (and, thus c) can just keep on growing indefinitely?

quit trying to sound smart you have no idea what you're talking about

1. prime factorize C
2. find the common factors of the powers of the prime factors.
3. each common factor will give you a pair

For example, say C = 1,866,240,000.
1. C = 2^12 × 3^6 × 5^4
2. The factors common to all of 12, 6, and 4 are only 1 and 2.
3. (2^12 × 3^6 × 5^4)^1 is one of the A^B pairs.
(2^6 × 3^3 × 5^2)^2 is the other.

No, it's NP-hard. What you're asking for fundamentally boils down tot he discrete logarithm problem.