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1967 IMO Problems/Problem 3

Revision as of 19:42, 6 September 2024 by Pf02 (talk | contribs)

Problem

Let $k, m, n$ be natural numbers such that $m+k+1$ is a prime greater than $n+1.$ Let $c_s=s(s+1).$ Prove that the product \[(c_{m+1}-c_k)(c_{m+2}-c_k)\cdots (c_{m+n}-c_k)\] is divisible by the product $c_1c_2\cdots c_n$.

Solution

For any $m, n$, $c_m - c_n = m(m+1) - n(n+1) = m^2 + m - n^2 - n = (m+n)(m-n) + m - n = (m+n+1)(m-n)$.

We can therefore write the product in the problem as follows:

\[(c_{m+1}-c_k)(c_{m+2}-c_k)\cdots (c_{m+n}-c_k)\] \[= \left(\prod_{a = 1}^n m+a+k+1\right)\left(\prod{a = 1}^n m+a-k\right).\]

But, the product of any $n$ consecutive integers is divisible by $n!$. We can prove this as follows:

\[(t-n+1)(t-n+2) \cdots (t) = \frac{(t+n)!}{t!} = n! \frac{(t+n)!}{t!n!} = n! {t+n\choose t}.\]

Therefore, $\prod_{a = 1}^n m+a-k$ is divisible by $n!$, and $(m+k+1)\left(\prod_{a = 1}^n m+a+k+1\right)$ is divisible by $(n+1)!$. However, we are told that $m+k+1$ is prime and therefore it is not divisible by any of the numbers $1$ through $n+1$. Therefore, $\prod_{a = 1}^n m+a+k+1$ is divisible by $(n+1)!$.

Finally, it is clear that $(c_{m+1}-c_k)(c_{m+2}-c_k)\cdots (c_{m+n}-c_k)$ is divisible by $n!(n+1)! = (1 \cdot 2)(2 \cdot 3)(3 \cdot 4) \cdots (n \cdot (n+1)) =$$c_1c_2\cdots c_n$. $\square$

~mathboy100

Solution 2

We have that $c_1c_2c_3...c_n=n!(n+1)!$

and we have that $c_a-c_b=a^2-b^2+a-b=(a-b)(a+b+1)$

So we have that $(c_{m+1}-c_k)(c_{m+2}-c_k)\ldots(c_{m+n}-c_k)=\frac{(m+n-k)!}{(m-n)!}\frac{(m+n+k+1)!}{(m+k+1)!}$ We have to show that:

$\frac{(c_{m+1}-c_k)(c_{m+2}-c_k)\ldots(c_{m+n}-c_k)}{n!(n+1)!}=\frac{(m+n-k)!}{(m-n)!n!}\frac{(m+n+k+1)!}{(m+k)!(n+1)!} \frac 1{m+k+1}$ is an integer

But $\frac{(m+n-k)!}{(m-n)!n!}=\binom {m+n-k}n$ is an integer and ${(m+n+k+1)!}{(m+k)!(n+1)!} \frac 1{m+k+1}=\binom {m+n+k+1}{n+1}\frac 1{m+k+1}$ is an integer because $m+k+1|m+n+k+1!$ but does not divide neither $n+1!$ nor $m+n!$ because $m+k+1$ is prime and it is greater than $n+1$ (given in the hypotesis) and $m+n$.

The above solution was posted and copyrighted by Simo_the_Wolf. The original thread can be found here: [1]


Remark and correction (added by pf02, September 2024)

1. The two solutions are essentially the same. They are just different expressions of the same idea. Unfortunately, both are written very sloppily, but the interested reader can easily correct the editorial mistakes and make sense of the proofs.

2. Both solutions are incomplete. They both assume (without saying it) that $m + 1 > k$. Maybe the problem meant to say this, but it didn't, so we have to see what happens if $m + 1 \le k$. There are two cases:

First, assume $m + 1 \le k$ and $m + n /ge k$. In this case, $(c_{m + 1} - c_k)(c_{m + 2} - c_k) \cdots (c_{m + n} - c_k) = 0$, and we can say that $0$ is divisible by anything.

Next, assume $m + n < k$. In this case, each factor of the first product is negative, and we will consider the product

$P = (c_k - c_{m + 1})(c_k - c_{m + 2}) \cdots (c_k - c_{m + n})$

for the sake of working with positive factors.




TO BE CONTINUED. I AM SAVING THIS SO I DON'T LOSE WORK DONE SO FAR.

See Also

1967 IMO (Problems) • Resources
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 4
All IMO Problems and Solutions
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