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I am stuck on proving a lot of the theorems that are discussed in number theory. Most of the theorems that I've seen in number theory so far are the ones I've already been shown how to prove, but I don't get the general approach for finding out how to prove them.

Let's say, for instance, we want to prove that there are infinitely many prime numbers. I know the basic methods of proving statements like direct proofs, proof by contradiction etc., so we could suppose that there are finitely many primes $p_1, p_2, \cdots,p_k$ with $p_1 < p_2 <\cdots < p_k$ by contradiction. Then in the next step of the proof, a new integer n is defined as $n = p_1\cdot p_2\cdot\space\cdots\space\cdot p_k + 1$, and because it is greater than $p_k$, it is composite since we assumed that there are finitely many primes. But the thing I don't get though is how $n = p_1\cdot p_2\cdot\space\cdots\space\cdot p_k + 1$ just came up so randomly.

I understand that continuing the proof eventually leads to the contradiction that 1 is composite, but the trouble is I don't know where to start just in general.

For example, if I wanted to prove that there are infinitely many primes of the form $6\cdot k + 5$ for some integer k, where would I start? I've tried doing something like defining $n = (6k_1 + 5)(6k_2 + 5)\cdots +(6k_r+5) + 5$ for $r$ primes of the form since I'm assuming that n could be a new prime of the form maybe (not too sure), but from there, nothing seems to be working out and I cannot get a contradiction in any way. And even if I did want to get a contradiction, how would I know what will end up being the contradiction in the end?

Is anybody able to help me with this? Thank you in advance.

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  • $\begingroup$ Just so you know you can accept someone's ( not mecessarily my) answer any time. $\endgroup$ – Roddy MacPhee May 4 at 14:05
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The problem is, proving isn't necessarily easy. It's historically, required inventing or reusing different parts of mathematics. There's not a general way to prove an unproven statement. There are general inference rules, but not a general proving method that work for all statements.

You could try proving, there's no natural number n above which all natural numbers are of form (6k+1)j-k with k,j greater than 0. But this method, is part of a restatement of the twin prime conjecture. This would prove infinitely many primes of form 6n+5 though, however impractically.

Simpler statements, and practicing with different principles, may help your proving ability.

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Proving almost anything in math can often involve dead ends. If you get to a certain level on this site, you will be allowed to see deleted answers.

In those deleted answers, you will see professionals and amateurs alike get close to the right answer and then realize that one little detail invalidates their whole argument.

Of course that wasn't an option for Andrew Wiles for his first attempt at a proof of the Fermat conjecture, and I certainly don't recommend you try to prove that one.

The theorem about infinitely many primes of the form $6k + 5$ (or $6k - 1$, if you prefer) seems a bit more manageable than the Fermat conjecture, now the Wiles theorem.

I would start by multiplying some arbitrary numbers of the form $6k + 5$ and then add 6. But the first difficulty I run into is that if I choose an even amount (what you call $r$) of numbers of the form $6k + 5$, their product is actually of the form $6k + 1$, e.g., $5 \times 11 = 55 = 6 \times 9 + 1$.

Next I would try either stipulating that $r$ has to be odd, or I would try multiplying that product by 6 and subtracting 1 from that, e.g., $6 \times 55 - 1 = 329 = 7 \times 47$. Hmm... this could actually work.

Even if it doesn't work, it helps me make the point that it is often very helpful to plug specific numbers into your equations, it can help you see things more quickly.

Maybe it would be better for you to try to prove even simpler things. Here's one for starters: given a prime $p$, prove that $p + 2$ or $p + 4$ may also be prime... but not both, with only one exception. The answer is on this site, but only look at that once you come up with the proof on your own.

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    $\begingroup$ There are two exceptions if you allow negative numbers... $\endgroup$ – David R. Mar 8 at 22:39
  • $\begingroup$ @DavidR. Right, yes, thank you, good catch. $\endgroup$ – Robert Soupe Mar 9 at 3:59
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First, the contradiction in the proof of infinitely many prime numbers is that this number $n=p_1p_2\dots p_k+1$ should be a multiple of a prime (fundamental theorem of artithmetic tells you that any number can be expressed as a product of primes), but by this construction, you can see that $n$ is not a multiple of any of these prime numbers (because when you divide $n$ by any of these prime numbers $p_i$ it gives you the residue $1$) which leads a contradiction. So, it is not random, the $n$ was constructed so that $n$ is not divisible by any of these prime numbers. Now, about primes of the form 6k+5 there may be methods similar to the above, but you may check Dirichlet's theorem of prime numbers in arithmetic progressions which is way more general https://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions The techniques to prove this are not elementary, instead it uses analytic properties of the L -functions associated to characters.

PS: So, for the case 6m+5: Note that besides 2,3 the primes are of the form 6m+1 or 6m+5. Assume by contradiction that $p_1,p_2,\dots, p_k$ are all the primes of the form 6m+5. Then, define $n=(2)(3)(p_1p_2\dots p_k)-1=6p_1\dots p_k-1$. The idea is that the primes $2,3,p_1,\dots,p_k$ do not divide $n$. Hence, the only primes that divide $n$ should be of the form $6m+1$, but if this were the case, then $n=(6m_1+1)(6m_2+1)\dots(6m_r+1)=6M+1$, so $n\equiv 1\mod (6)$, but this is a contradiction since by construction $n=6p_1\dots p_k-1\equiv -1\mod (6)$

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I think that there is no proper method although you can point out some necessary but not sufficient rules to solve a mathematical problem (prove a theorem is one of these). First, you have to fully understand the statement of the problem. Second, we must take into account the mathematical knowledge that could help find the solution (the more you have, the better). You need to be lucid to perceive the best possible way for which there are usually several, for example, the famous theorem of quadratic reciprocity (which is a difficult problem!) has several different proofs.

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  • $\begingroup$ And restatements can help and hurt. There are infinitely many restatements of Fermat's little theorem for example. $\endgroup$ – Roddy MacPhee Mar 9 at 0:41

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