TT's Maths Thread

[TrueTears]

Notice that B is the middle number (again showing some nice symmetry ). So yeah, actually just do your big general approach, works best, and then split it into a product into many little products that involve only the single prime, and then just prove that for each of these products, you get the 'middle number' both for lcm and gcd. and yeah result easily follows from here. (this is ussually a good approach, to focus on single primes and then take the product of all the singles).

I don't really get this last part, what do you mean smaller products? what product? Can you show me?

Thanks

Actually dw I get it now

[TrueTears]

A few more questions

1. Show that ends with zeros, generalise as well.

2. Prove that there are infinitely many primes of the form where , ie, this sequence .

3. Prove that among any consecutive positive integers there is at least one which is smaller than the sum of its proper divisors. (The proper divisors of a positive integer are all positive integers other than and which divide . For example, the proper divisors of are and .)

4. Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression.

[kamil9876]

nice ones

Here are some clues

1.) basically this is equivalent to proving that there are exactly 249  5's in the product . All multiples of 5 contribue at least one 5. multiples of 25 contribue two 5's, multiples of 125 contribue three 5's, multiples of 625 contribue four 5's....

2.) This problem can be done very efficiently with modular arithmetic, but without it note that any odd number of the form 4k+3 has at least one prime factor of the form 4k+3; for if it didn't then it would have it's prime factors all of the form 4k+1 and (4a+1)(4b+1)(4c+1).... would be of the form 4k+1 if you expand out.

Clue number 2: Euclid's proof on infinitude of primes.

3.) try a stronger result: all multiples of 12 have this property. If you can prove this then it implies this result since amongst any 12 consecutive numbers we have one multiple of 12.

4.) This is actually more of an algebra problem, my solution is pretty algebraic. suppose they were in an arithmetic progression of length d:

for some k

for some m

therefore:



now you can rearrange this into an equation of the form:



where a,b,c are rational. and the cube roots are obviously irrational and distinct. Prove that this cannot be the case (algebra exercise).

[brightsky]

wow...good job.

[TrueTears]

1. Show that ends with zeros, generalise as well.

Let's count the number of 's in the sequence .



Let's count the number of 's in the sequence .



By comparing summands we see that there exists much more 's in the total amount of PPF's of every number in the sequence

Now that means there exists zeros!

[kamil9876]

2. Prove that there are infinitely many primes of the form 4k+3

2.) This problem can be done very efficiently with modular arithmetic, but without it note that any odd number of the form 4k+3 has at least one prime factor of the form 4k+3; for if it didn't then it would have it's prime factors all of the form 4k+1 and (4a+1)(4b+1)(4c+1).... would be of the form 4k+1 if you expand it out

ok here is the completion:

assume there are only finitely many primes of this form. Take the product of all of these, call this M. M will either be of the form 4k+1 or 4k+3:

case1: M is the form 4k+1

then M+2 is of the form 4k+3 and hence by the lemma in the second quote above: it must be divisible by some prime of the form 4k+3. Let 4a+3 by such a prime, then is not an integer since M is divisible by 4a+3. But this is true for all 4a+3 hence there must be some bigger one.

case2: M is of the form 4k+3

then M+4 is of the form 4k+3, and we can do a similair proof as in Case1.


Some interesting history:

Fermat proposed this sort of problem (perhaps even the same 1). to prove for 4k+1 it is much more difficult. Euler conjectured that there are infinitely many primes of the form 1+ak for all . Dirichlet later proved something in fact even stronger, that there are infinitely many primes in any arithmetic sequence where the gcd of the common difference and initial value is 1. see more here

[TrueTears]

2. Prove that there are infinitely many primes of the form where , ie, this sequence .

Hehe, pretty similar to Euclid's proof

Assume there are finitely many primes where is the largest prime in this sequence.

Now consider a number

is either prime or composite.

can not be prime because it would contradict the fact that is the largest prime.

However we know that must have at least prime divisor, , by the Fundamental Theorem of Arithmetic.

However this prime can not be any number in the sequence since that will result in a remainder of .

Thus this contradicts our assumption that there are finitely many primes in this sequence, therefore there must be infinite many primes in this sequence.

[kamil9876]

but how do u know p is of the form 4k+3? could be of the form 4k+1...

[TrueTears]

note that any odd number of the form 4k+3 has at least one prime factor of the form 4k+3

How do you prove that?

[kamil9876]

suppose there was a number of the form 4k+3 that had no prime factor of them form 4k+3. Hence it has prime factors all of the form 4k+1. and so it can be written as:

(4a+1)(4b+1)....

which when expanded out has a +1 at the end, and all otehr terms multiples of 4. Hence it is of the form 4k+1, a contradiction.

[TrueTears]

But if you're talking about the PPF of 4k+3 where's the exponents?

[kamil9876]

you can write then in if u want but still u can write 3^2 * 7 as 3*3*7 so i guess u could say it is (4*0 + 3)(4*0 + 3)(4*1 + 3). ie a,b,c... etc do not have to be distinct. But if u really want just add in the exponents if u want and u still get same result.

[TrueTears]

Ah ok, I get it now.

[TrueTears]

3. Prove that among any consecutive positive integers there is at least one which is smaller than the sum of its proper divisors. (The proper divisors of a positive integer are all positive integers other than and which divide . For example, the proper divisors of are and .)

Let's list the first consecutive positive integers.



The proper divisors of each are as follows:

























So only satisfies the condition as

So if we can find every multiple of then it must also satisfy this condition.

Every consecutive positive integers must contain a multiple of .

Let this multiple of be where

Then the proper divisors would contain at least

Thus


I have a few questions to this question though, why does only multiples of 12 work? (I just tried by experimentation and 12 worked, but why does it work?)

Also the is not the real sum of the all the proper divisors, it just includes some, eg when we have

But we are missing the proper divisors and . So the real sum of the divisors should be rather than , but is already larger than so the condition is satisfied...

But don't I need to somehow include the sum of ALL possible divisors in my generalization or is it alright to just prove that if some divisors' sum exceeds then I've proven what I needed?

[kamil9876]

yeah. let by that logic, a generalisation would be sd(ab)> a * sd(b). where sd is sum of divisors.