(Dis)Proving a Math Conjecture

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(Dis)Proving a Math Conjecture

Postby Baltorigamist » July 16th, 2018, 3:26 am

Something occurred to me today, but I don't know how to prove or disprove it. Since some of you are a lot better at math than me, I figured I'd post it here.

My conjecture is this:
Let n be the product of any two prime numbers a and b. Then n has only four factors: 1, n, a, and b.

Let me know of any proofs (or counterexamples).

Thanks.
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Re: (Dis)Proving a Math Conjecture

Postby Kabuntan » July 16th, 2018, 6:05 am

I can't prove it either, but for me, your conjecture is true.
Because:
n = a * b
and a and b are prime factors (i.e. they can't be written as a product of two or more factors bigger than 1),
n can not be written as a product containing more than two bigger-than-1 factors (which are a and b).

I guess that's not a valid proof though.
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Re: (Dis)Proving a Math Conjecture

Postby origami_8 » July 16th, 2018, 11:45 am

It has to be true because if you make a prime factorization, you would come back to a and b.
Every product n can be diverted by 1 and itself, so those are clear too.

If you take more than two prime numbers you can have more factors because you can have multiple prime numbers counted together in different ways. Lets take the number 12 for example. It is made up of the prime numbers 2,2,3 so you could see it as 2x2x3 or as 4x3 or as 6x2, but the prime numbers a number is made up of, stay always the same. There is no way 5, 7 or 11 could ever be a factor of 12 it is always 2, 2 and 3. The four and six are only possible because they are made up of the same prime numbers as 12. 4=2x2 leaving a 3 left to get to 12, and 6=3x2 leaving a 2 left to get to 12.
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Re: (Dis)Proving a Math Conjecture

Postby zurathan » July 16th, 2018, 11:01 pm

Baltorigamist, I may have no proof or disproof but in the end this may not even need one. In fact I'm counting on that probability. A litlebit more understanding is enough imho.
lets just think n=a^1*b^1 not just a*b
the definition and divider number is 2*2, each 2 is previous povers plus one. cuz any number pover zero is 1 and incluced.
first divider is a^0*b^0, 2nd is a^1*b^0,3rd a^0*b^1, 4th a^1*b^1.

for example if n=a^c*b^d, then divider number is (c+1)*(d+1), both "a" and "b" must be prime ofcourse... every power combination including zeroes are divider

this may help, or not... stay curious. excuse my english.
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Re: (Dis)Proving a Math Conjecture

Postby Goldtiger-997 » July 17th, 2018, 1:26 am

Here's a more "rigorous" proof:

Assume for a contradiction that n has another factor, c ≠ 1, a, b, n.
Hence c divides n = ab
The only factors of a are 1 and a, and the only factors of b are 1 and b
Therefore the only factors of ab are 1, a, b, ab
Which implies c is one of 1, a, b, n (because n = ab)
But at the start we said that c ≠ 1, a, b, n.
Contradiction.
Therefore there do no exist any numbers other than 1, a, b, and n that are factors of n.
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Re: (Dis)Proving a Math Conjecture

Postby Baltorigamist » August 15th, 2018, 7:27 pm

Thanks, everyone!
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Re: (Dis)Proving a Math Conjecture

Postby chesslo » July 9th, 2019, 12:27 pm

If you denote d(n) to be the number of divisors of n then it is easy to see that the function d is a multiplicative function of n. In particular, if n is a product of two distinct primes a,b then d(n)=d(ab)=d(a)d(b)=2*2=4. If n is a square of a prime then d(n)=3. If you are interested in elementary number theory, I recommend the book: Elementary number theory by David Burton.
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