Difficult number theory problem proofs

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The discussion revolves around a challenging number theory problem that involves proving the inequality p1p2...pt < 4^n for primes not exceeding a positive integer n. A proposed approach involves using the natural logarithm and the prime number theorem to establish a framework for the proof, suggesting that the left side approximates the nth prime. However, participants note that while the approach hints at the inequality's validity for large numbers, it does not constitute a formal proof. There is a consensus that additional constraints on prime density and checks for smaller numbers are necessary for a complete proof. The conversation also touches on the broader topic of the difficulty of proving certain prime number conjectures.
pat devine
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The following is a repost from 2008 from someone else as there was no solution offered or provided I thought id post one here
Homework Statement neither my professor nor my TA could figure this out. so they are offering fat extra credit for the following problem

Let n be a positive integer greater than 1 and let p1,p2,...,pt be the primes not exceeding n.

show that p1p2...pt<4^n- An mathematically rigourous proof of this would take a long time but I would suggest the following outline which is logically correct I believe:

First we need to consider a lowest bound value of n and for that purpose is can be stated as pt above

so we want to show that p1*p2*p3...*pt < 4^pt

consider the natural log of both sides
log (p1*p2...) = log(4^pt)
consider the left hand side
log(p1*p2*p3...) = log p1+ logp2 + log p3 ...

Now consider the prime number theorem which can be restated such that:
logp1+logp2+logp3...+logpt ~ pt

another way to think about using a well known restatement of the prime number theorem is that the density of primes at any point n (ie the approximate prime gap) is approximately equal to log n.

eg logp1 is ~ the gap between p1 and p2
log22 is ~ the gap between p2 and p3 and so on

therefore is you sum up all the gaps between the first n primes the sum will be (approx.) the value of the nth prime pn.
Therefore the Left hand side = pt (approx within the order +/- 1% for large p)
Now the right hand side log(4^pt) = pt* log 4 = 1.386 * pt
Therefore this proves the original statement since pt < 1.386 * pt

Any comments would be welcome

Pat
 
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(I removed the boldface and moved the thread to the homework section)

That approach suggests that the inequality is true for large numbers, but it is not a mathematical proof. Note that locally the ratio between the two sides can decrease, e.g. between 101 and 103 (both are prime, and 103 > 16).
You'll need some upper limit on the density of primes, probably together with a manual check of the smaller numbers (this part is unproblematic).
 
Hello and thanks for that. I know that it's not a a proof but I think it provides a framework to achieve one. In any event it's nothing of any value. I agree with your suggestions however. I guess I was trying to find some proveable number theory problems and found this one. I suspect without basis that many prime number conjectures are true but not proveable. If you have any difficult but proveable suggestions I'd like to hear them.
Many thanks for replying it's my first day here and it's good to get feedback.
P
 
I'm sure it is possible to prove that inequality.

I would expect the same inequality to hold for 3n.
I guess for every number w>e there are at most a finite set of elements where the product exceeds wn. Which also means there is (a) no product of primes that exceeds en or (b) there is a largest w>e where the inequality is violated somewhere.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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