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Homework Help: Prime factors of binomials

  1. Dec 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Is it true that for each ##n\geq 2## there are two primes ##p, q \neq 1## that divide every ##\binom{n}{k}## for ##1\leq k\leq n-1##?


    Examples:

    For ##n=6: \binom{6}{1}=6; \binom{6}{2}=15; \binom{6}{3}=20; \binom{6}{4}=15; \binom{6}{5}=6.## So we can have ##p=2## and ##q=3##.


    For ##n=3: \binom{3}{1}=3; \binom{3}{2}=3.## So we can have ##p=3## and ##q=17## (or any other prime). We just have to show that each ##\binom{n}{k}## is divisible by at most two primes.


    2. Relevant equations
    ##\binom{n}{k}=\frac{n!}{k!(n-k)!}##

    3. The attempt at a solution
    Proof by induction:
    Base case: True for ##2, 3##
    Induction on ##n##: Assume true for ##n##. Prove for ##n+1##:
    $$\binom{n+1}{k}=\frac{(n+1)!}{k!(n-k+1)!}=\frac{n+1}{n-k+1}*\frac{n!}{k!(n-k)!}$$

    Now what? Is it possible to say that since ##\frac{n!}{k!(n-k)!}## is divisible by primes (by assumption) that somehow ##\frac{n+1}{n-k+1}*\frac{n!}{k!(n-k)!}## is as well?
     
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  3. Dec 11, 2014 #2

    ShayanJ

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    [itex]p|b \Rightarrow b=pk \Rightarrow ab=pak \Rightarrow ab=pk' \Rightarrow p|ab [/itex].
     
  4. Dec 11, 2014 #3
    Can you elaborate on your solution? I can't seem to understand it very well.
     
  5. Dec 11, 2014 #4

    haruspex

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    That is not going to work. The new divisor (n-k+1) might remove the factor of interest.
    It's not clear to me how induction is going to succeed. You are not trying to show the same two primes work for all n.
    That's not what was originally stated. It reads more like 'at least two distinct primes'. But that is not true (as your n=3 example shows). So I guess the problem statement should have been:
    Is it true that for each n≥2 there are at most two distinct primes p,q≠1 that divide every ##\binom{n}{k}## for 1≤kn−1?​
    Please confirm.
     
  6. Dec 11, 2014 #5

    mfb

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    The problem statement is a bit misleading. For every ##n \choose k##, just either p or q has to be a factor?

    A proof by induction... I'm not so convinced (unless you choose a very clever way of induction).
    What happens if n is a prime or a prime power?
     
  7. Dec 11, 2014 #6

    haruspex

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    Doesn't sound to me that it's supposed to be the same p and q for each n. Did you read my post #4?
     
  8. Dec 11, 2014 #7
    Sorry. I don't know why I put the "at most" in there. I want to prove that two primes ##p, q## exist that can divide ##\binom{n}{k}##. Sometimes only 1 prime is needed (i.e. when ##n## is prime). However, I want to prove that with only two primes, I can divide ##\binom{n}{k}##.

    Also, each ##n## will have different primes ##p, q##.

    Please ask if I am not clear anywhere else.
     
  9. Dec 11, 2014 #8

    haruspex

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    In what sense 'needed'?
    I cannot guess what you mean by that.
     
  10. Dec 12, 2014 #9

    mfb

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    After I posted, yes (was doing some checks before, didn't see your post earlier).

    I think post 7 confirms my interpretation: "For a given n, there is a set {p,q} of two primes such that every ##n \choose k## (apart from k=0 and k=n) has p or q as a factor."
     
  11. Dec 12, 2014 #10
    ##n## prime is a simple case, as is ##n-1## prime.

    You might be able to to run induction on multiples of each prime, perhaps; I haven't tried it though.
     
  12. Dec 12, 2014 #11

    PeroK

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    It might be that if p / n and q / (n - 1) with p not = q then p and q suffice.
     
  13. Dec 12, 2014 #12
    Your bonus this week, courtesy of number theory, is that you get p≠q thrown in free! :)

    If I look at n=15, it depends how you choose {p,q}: {5,7} and {3,7} work, {5,2} and {3,2} don't.
     
  14. Dec 12, 2014 #13

    mfb

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    At n=15, all numbers are divisible by 5 and 3. I think that is not a coincidence (tested up to 39, Excel gets troubles with large numbers afterwards)
     
    Last edited: Dec 12, 2014
  15. Dec 12, 2014 #14

    PeroK

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    That's because pq < n/2. Proof is slippery but if n fails then pq < n/2 where p and q are largest prime divisors of n and n-1.

    Also, n can't have a repeated prime factor.
     
  16. Dec 12, 2014 #15

    mfb

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    5 and 7 are the largest prime divisors of 15 and 14, but 5*7 is not smaller than 15/2.
    Also, which proof?
     
  17. Dec 12, 2014 #16

    PeroK

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    I thought I was closing in on a proof but I'm not so sure now. If you could find n and n-1 where both are the product of distinct primes with pq < n/2 then I think the result might fail for that n.

    The key point is cancelling factors. I think every binomial is divisible by p or q until k = pq. Where p and q are any prime factors of n and n-1. In particular if the result fails for n, then pq must be < n/2 when they are the largest prime factors.

    There might be a counterexample but hard to find.
     
  18. Dec 12, 2014 #17

    haruspex

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    Suppose prime p divides n, r times over. Characterise those k for which p is not a factor of ##\binom n k##.
     
  19. Dec 12, 2014 #18

    mfb

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    @PeroK: n and n-1 are always the product of distinct primes. Any shared factor would also be a factor of their difference, which is 1.

    63 has 7 as largest prime factor, 64 has 2, and 7*2=14 < 64/2.
    64 has 2 as largest prime factor, 65 has 13, and 2*13=26 < 65/2.

    Those are the smallest examples. 80 and 81 is the next pair and does not have 2 as one of the primes. Those three are the only examples below 100 and I don't want to check more now. Hypothesis: there is an infinite number of examples.
     
    Last edited: Dec 12, 2014
  20. Dec 12, 2014 #19

    PeroK

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    Both n and n-1 must be products of distinct primes I.e no repeated factors. Repeated factors won't completely cancel.

    You need something like:

    n = 3.7.11...

    n -1 = 2.5.13...
     
  21. Dec 12, 2014 #20

    haruspex

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    Hint 2: Lemma: k must be a multiple of p,.
     
  22. Dec 12, 2014 #21
    Up to 110, it's possible to find two prime factors of all non-prime-power n for which one or the other divide every C(n,k) in range. 110 doesn't work that way, although it is possible to find two other primes that work (109 being the easy choice). This might help set the direction of any attempted proof.

    2 does not divide C(110,10) and C(110,44) (among others)
    5 does not divide C(110,10) and C(110,55) (among others)
    11 does not divide C(110,44) and C(110,55) (among others)
     
  23. Dec 13, 2014 #22

    PeroK

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    Here's my idea for a counterexample:

    Suppose you could find n such that:

    ##n = p_1p_2...p_r##
    ##n-1 = q_1q_2...q_s##

    Where the product of any two of these primes is < n/2. Then, for any two of these primes, we have a k, such that neither prime divides ##binom{n}{k}##

    Any such n will be quite large: the largest primes ##p_r## and ##q_s## must be at least 20, so n/2 must be at least 400.

    In any case, if you are going to prove the proposition, then I think you will have to show that no such n can exist.

    My guess is that the proposition fails, although finding a countereaxmple will be hard.

    PS I had a scout round online. There are a lot of consecutive squarefree integers, but I couldn't find anything relevant about the size of their prime factors.
     
    Last edited: Dec 13, 2014
  24. Dec 13, 2014 #23

    mfb

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    I found a counterexample:
    6761*6823*6827 = 314,931,578,581
    2*37*827*5146109 = 314,931,578,582
    6827*5146109 = 35,132,486,143

    As you can guess from the first number, I just tested the product of three primes in that range to have a square-free number without a very large prime factor next to it. As I was successful after ~20 attempts, this is almost certainly not the smallest pair.
     
  25. Dec 13, 2014 #24

    haruspex

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    Let ##n = \Pi p_i^{r_i}## and ##f = \max \{p_i^{r_i}\}##. Similarly, let g be the largest prime power factor of ##\{n-f, n-f+1, .. , n-1, n/f\}##. I believe n is a counterexample if and only if n > fg. Will attempt to test this.
     
  26. Dec 13, 2014 #25

    PeroK

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    Yes, I forgot I was trying to find a solution using prime factors of n and n-1 only. So, my idea of a counterexample was only for that!

    I can't see that there could be a counterexample with your constraint. There's bound to be a large prime factor somewhere in that list.
     
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