2013 Indonesia MO Problems/Problem 6

Problem

Suppose $p > 3$ is a prime number and $S = \sum_{2 \le i < j < k \le p-1} ijk$ Prove that $S+1$ is divisible by $p$ .

Solution

if you let $j$ be constant, you can think of it as summing for each $i$ , $\sum^{p-1}_{k=j+1}$ , and since its for all $i$ you can add another sum to get $\sum^{j-1}_{i=2}\sum^{p-1}_{k=j+1}$ , and for all $j$ we can add another sum, to get $\sum_{2\leq i<j<k\le p-1}ijk=\sum^{p-2}_{j=2}\sum^{j-1}_{i=2}\sum^{p-1}_{k=j+1}ijk$ $\sum^{p-2}_{j=2}\sum^{j-1}_{i=2}\sum^{p-1}_{k=j+1}ijk$ $\sum^{p-2}_{j=2}\sum^{j-1}_{i=2}ij\sum^{p-1}_{k=j+1}k$ $\sum^{p-2}_{j=2}\sum^{j-1}_{i=2}ij\left(\frac{(p-j-1)(p+j)}{2}\right)$ $\frac{1}{2}\sum^{p-2}_{j=2}\sum^{j-1}_{i=2}i(j(p-j-1)(p+j))$ $\frac{1}{2}\sum^{p-2}_{j=2}\frac{(j+1)(j-2)(j(p-j-1)(p+j))}{2}$ $\frac{1}{4}\sum^{p-2}_{j=2}(j+1)(j-2)(j(p-j-1)(p+j))$ $\frac{1}{4}\sum^{p-2}_{j=2}-j^5+j^3p^2-j^3p+3j^3-j^2p^2+j^2p+2j^2-2jp^2+2jp$ $\frac{1}{4}(\sum^{p-2}_{j=2}-j^5+(p^2-p+3)\sum^{p-2}_{j=2}j^3+(-p^2+p+2)\sum^{p-2}_{j=2}j^2-(2p^2-2p)\sum^{p-2}_{j=2}j)$ $\frac{1}{4}(-\frac{2n^6+6n^5+5n^4-n^2}{12}+(p^2-p+3)\frac{(p-2)^2(p-1)^2}{4}-(p^2-p+3)+(-p^2+p+2)\frac{(p-2)(p-1)(2p-1)}{6}+(p^2-p-2)-(2p^2-2p)\frac{(p-1)(p-2)}{2}+(2p^2-2p)$ $\frac{1}{48}(p^6-7p^5+11p^4+3p^3+12p^2-20p-48)=S$ $S+1=\frac{1}{48}(p^6-7p^5+11p^4+3p^3+12p^2-20p-48)+1=\frac{p}{48}(p^5-7p^4+11p^3+3p^2+12p-20)$ since it has a factor pf $p$ , we need to prove $\frac{p^5-7p^4+11p^3+3p^2+12p-20}{48}$ is always an integer, for prime $p>3$ $p\mod 6\equiv 1\text{ or }5$ , so let $p=6k+1$ and $p=6k+5$ for the first case, you get $\frac{24k(324k^4+108k^3-63k^2+6k+7)}{48}$ and for the second you get $\frac{24(3k+2)(108k^4+252k^3+195k^2+54k+5)}{48}$ , notice how both of these are always integers, thus it is proven $p$ divides $S+1$

2013 Indonesia MO (Problems)
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8	Followed by Problem 5
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2013 Indonesia MO Problems/Problem 6

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