Number Theory

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Here is an example problem I am doing at the moment - this is supposed to be extremely easy, but I don't see how to go about working on it.

NB: Please don't spoil it for me, I am just looking for a rough idea of what kind of thing I should be doing to crack these kinds of questions.

Whether you can solve number theory problems or not is determined by your genes. If you have to ask this question, the answer is pretty much clear.

>>7776746
>
Nice bait, senpai.

>>7776754

Wh-what? Am I missing something obvious?

I don't study maths, its an extracurricular thing I am trying to learn in my own free time.

>>7776746
With this problem in particular, it might be a good idea to check a few examples and see if it will hold. Does this work for 4? 7?
If you can find an example where it doesn't hold then it obviously doesn't hold for all cases.

>>7776746
It's clearly false:
>7 is prime
>Thus has no factors other than itself and 1
>1 isn't prime
>Thus 7 can not be written as a product of prime factors.

QED

>>7776762

4 = 4 (prime)
7 = 7 (prime)
10 = 10(prime)
13 = 13(prime)
16 = 4 × 4
19 = 19(prime)
22 = 22(prime)
25 = 25(prime)
28 = 4 × 7
etc etc.

But i dont think this is a good method, because even if it works then it is just dumb luck - i had no way of knowing the answer would be easy, for all I know it works fine for the first 9 million numbers before only going wrong then.

>>7776776

No, you have a new definition of "prime" according to the set - eg: 4 is the new lowest "prime", followed by 7 then 10, etc etc.

>>7776782
10*10 = 4*25

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103KB, 1000x968px

>>7776816

i have mixed feelings

on the one hand, now i know the answer
on the other, you ruined it for me.

oh well. Spilt milk and all that. Could you tell me the general method you would use to solve a problem of this type?

>>7776822
The explanation may be more satisfying. Since you have "primes" that aren't actually primes, there is nothing to stop these "primes" from being made up of real primes that can be rearranged to form new "primes". So let's pick two pairs of primes that can be rearranged.

We might start with the lowest primes 2*2 and 3*3 which can be rearranged as 2*3 and 2*3, but we must then notice that this cannot work because 3k+1 can never have 3 as a factor. So let's get rid of the 3s and replace them with 5s:

2*2 * 5*5 = 2*5 * 2*5

And voila, you have a counterexample. The reason set S does not have the unique prime behavior is quite simply that the "primes" aren't primes.

>>7776840

Okay so I wanted to try this with a different series so I just made up one where the values = 3k - 1

So the set goes {2,5,8,11,14,17,20,.... etc}

Now some of the first few primes, 2 5 and 11, are already in the series, but 3,7,13,19.... arent.

Now we cant use the 3, as 3k-1 will never be divisible by three. So ive tried 7x7x13x13, which is 8281. this cant be right because 8281+1 is not divisible by three.

I tried next 7x7x19x19, 13x13x19x19, 19x19x31x31, 13x13x31x31 and 7x7x31x31, none of those answers fit because (ans + 1) is not divisble by three. Now obviously I am just doing trial and error which is hopeless. So I think maybe I misunderstand your explanation?

Artin reciprocity, UF/Dedekind domains, scheme theory, extension fields, Diophantine approximation, Galois representations, etc.

>>7776746
I was going to suggest that you consider the case where k =11. If k = 11 then s = 34. 34 has a prime factorization of 2*17. 2 is not contained in s, and 17 is not contained in s, meaning that 34, a member of s, cannot be expressed as the product of two other members of s. Therefore not all numbers in s can be expressed as products of other numbers in S.
But >>7776776 seems to have a better proof.

Prime factorization, Bezout's theorem, the pigeonhole principle and euclidean division (and modular arithmetic) are the most basic tools and already allow you to do a lot of stuff

>>7776906

>pigeonhole principle

T H I S

easily the most useful thing for these kinds of things

>>7776898

i think in the question you consider 34 as a pseudo-prime just for set S

>>7776782
In what universe is 4 prime?

>>7776880
The pairs don't have to be the same number.

2*2*2*5 = 2*20 = 8*5

>>7776822
>>7776782
This is wrong, since the definition of prime is that its only divisors are 1 and itself, and 1 is not in the set, so 4 is not the product of anything,since the least product is 4*4=16, so the set to begin with is flawed.

>>7776822
These types of problems can be solved by considering an arbitrary product between to members in the set [math]3k+1, 3m+1[/math]. Check to see if their product creates a new element in the set, ie:
[eqn](3k+1)(3m+1)=9km+3k+3m+1=3(3km+k+m)+1=3n+1[/eqn]
So you can see that the product of any two numbers in that form create another in that form

>>7776782
>>7776921
Sorry shouldn't have been so quick to judge, but can you explain why the numbers you have listed are prime for the set? I'm sure the answer is obvious.

>>7776927
reformatting the latex:
[math](3k+1)(3m+1)=9km+3k+3m+1=3(3km+k+m)+1=3n+1[/math]

>>7776743
there is no general purpose toolkit for number theory
just a bag of tricks that you pick up as you solve more

>>7776927
>So you can see that the product of any two numbers in that form create another in that form
That doesn't actually help answering the question though. The question is essentially asking if there exists (a,b) =! (c,d) such that 3ab+a+b = 3cd+c+d

>>7776928

So the question is poorly phrased I guess.

Try this:

Is it true that there is only one way to express numbers in set S as products of the smallest numbers possible in set S disregarding order?

eg like how prime factorisation works, but with set S instead of set {primes} as your base.

>>7776944
Ok I get it now thank you.

>>7776743
Start with basic proofs and go from there. Understanding of primes is a must.