# Prime factors of binomials

## Homework Statement

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.

## Homework Equations

##\binom{n}{k}=\frac{n!}{k!(n-k)!}##

## 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?

ShayanJ
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$p|b \Rightarrow b=pk \Rightarrow ab=pak \Rightarrow ab=pk' \Rightarrow p|ab$.

Can you elaborate on your solution? I can't seem to understand it very well.

haruspex
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$p|b \Rightarrow b=pk \Rightarrow ab=pak \Rightarrow ab=pk' \Rightarrow p|ab$.
That is not going to work. The new divisor (n-k+1) might remove the factor of interest.
Proof by induction:
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.
We just have to show that each (nk)\binom{n}{k} is divisible by at most two primes.
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?​

mfb
Mentor
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?

haruspex
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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?
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?

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##.

haruspex
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Sometimes only 1 prime is needed
In what sense 'needed'?
I want to prove that with only two primes, I can divide ##\binom{n}{k}##
I cannot guess what you mean by that.

mfb
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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?
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."

• Joffan
##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.

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.

It might be that if p / n and q / (n - 1) with p not = q then p and q suffice.

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.

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)

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PeroK
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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.

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.

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?

PeroK
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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?

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.

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##.

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.

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PeroK
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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.
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...

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##.
Hint 2: Lemma: k must be a multiple of p,.

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)

• mfb
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.

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mfb
Mentor
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.

• PeroK
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.

PeroK
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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.

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.

haruspex
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There's bound to be a large prime factor somewhere in that list.
It only needs to be a power of a prime, but it's still not obvious there'll be one. E.g. for n = 2520, f = 9. Doesn't leave that many candidates for the contributor of g.

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.
I tried all n up to a million. The furthest I had to search through the sequence n-1, n-2, ... for the second prime power was 19 (n = 520260, f = 29, g = 520241).

mfb
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I lost the connection of this f,g and n, n-1 thing to the original problem.

PeroK
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The original problem was to find common prime factors of all the binomials. So, for n, we had to find two primes p & q, at least one of which divides each binomial ##\binom{n}{k}##.

I tried to prove that picking the biggest prime factor of n and n-1 respectively would do. But, this didn't work, and you found a counterexample.

Certainly one prime must be a factor of n. But, you don't have to pick the second prime from n-1. If you choose f as the biggest prime factor of n (as defined above, including powers of the factor). Let's say ##f = p^r##, then p will divide all the binomials up to but not including ##k = p^r##.

So, you must choose your second prime factor from ##n-1, n-2 ... n-p^r##

To find a counterexample, you need not two consecutive numbers with "small" prime factors but a larger set of consecutive numbers that all have exclusively small enough prime factors.

• mfb
PeroK