Difference between revisions of "1967 IMO Problems/Problem 3"

Revision as of 23:07, 12 December 2022

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

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]

@@ Line 3: / Line 3: @@
 ==Solution==
-We have that <math>c_1c_2c_3...c_n=n!(n+1)</math>
+We have that <math>c_1c_2c_3...c_n=n!(n+1)!</math>
 and we have that <math>c_a-c_b=a^2-b^2+a-b=(a-b)(a+b+1)</math>

1967 IMO (Problems) • Resources
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Difference between revisions of "1967 IMO Problems/Problem 3"

Revision as of 23:07, 12 December 2022

Problem

Solution

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