http://www.kalva.demon.co.uk/imo/isoln/isoln786.html Alternate (killing) solution : So, in this probem, we're asked to prove that any partition $\{P_1, P_2, \ldots, P_6\}$ of the set $\{1, 2, \ldots, 1978\}$ is such that there exists $i \in \{1, 2, 3, 4, 5, 6\}$, and numbers $a, b, c \in P_i$ such that $a + b = c$. But, by a well-known result (for any integer $k \ge 2$, $R_k \le \lfloor k!e\rfloor + 1$), we have : \[ R_6 \leq \lfloor 6! \times e \rfloor + 1 = 1958, \] so, we can even replace 1978 with 1957.

mathmanman wrote: http://www.kalva.demon.co.uk/imo/isoln/isoln786.html Repaired link: http://mks.mff.cuni.cz/kalva/

Consider a complete graph $K_{1979}$ with $1979$ vertices numbered as $1, \ldots, 1979$. Consider the vertices $a,b \in K_{1979}$, if the number $|a-b| $ is from country $c_i$, color the edge joining them by color $c_i$. By Ramsey number theorem, on $6$ coloring the graph $K_{1979}$, there will be a monochromatic sub graph with $3$ edges. Since $1979 > \lfloor {e\cdot 6!} \rfloor + 1 \ge R_6(3)$. So let the edges joining the vertices $a,b,c$ of $K_{1979}$ have same color. WLOG $a>b>c$. Then $a-b , b-c, a-c$ are from the same country and $a-b + b-c = a-c$, so we are done.

Can someone expand on this? I don't understand what $R_k \le \lfloor k!e\rfloor + 1$ signifies and how it relates to the problem.

I think Learner94 has already explained it pretty well. Anyway, here $R_k$ denotes the Ramsey number $R(3, 3, \ldots, 3),$ with $k$ $3,$ and the IMO problem asks for a proof of Schur's theorem, with a bound on the cardinality of the set which is to be $k$-coloured (in the problem, $k = 6$). You might also want to read the following thread (and follow the links therein): Partition N s.t. a1 + a2 in B and b1*b2 in A The "twice as large as the number of one member from his own country" in the statement comes from the 'issue' raised in the above thread.

I'm wondering how we get the bound $R_k<k!e$. Here is my attempt to prove that. There is a mistake somewhere but I can't see where. We attempt to partition $S={1,...,n}$ into $k$ disjoint sets each of which is sum-free. One of the $k$ sets $A$ contains at least $n/k$ elements of $S$. The elements of $A$ generate at least $n/k-1$ distinct differences, so there are $n/k-1$ elements of $S$ not lying in $A$. Thus one of the remaining $k-1$ sets $B$ has at least $(n/k-1)/k$ numbers, which generates at least $(n/k-1)/k--1$ differences, which can't lie in $A$ or $B$. Continuing like this $k-1$ times, we end up with at least $(n-1)/k!-2!/k!-...-k!/k!$ numbers which can't lie in any of the sets. If this number is at least $1$, then any partition of $S$ into $k$ subsets contains a set which is not sum free. This suggests $R_k \le n <k!/(1!+...+k!)$. We know $1/0!+1/1!+1/2!+...+1/k!$ tends to $e$, so I am close but I obviously made a silly mistake somewhere.

my solution let $\{S_1,....,S_6\}$ partition of the set $\{1,2,...,1978\}$ let $S_i=\{a_1<a_2<...<a_{|S_i|}\}$ and $S'_i=\{a_1<a_2<...<a_{|S_i|-1}\}$ and $T_i=\{a_{|S_i|}-a_1>a_{|S_i|}-a_2>...>a_{|S_i|}-a_{|S_i|-1}\}$for$i=1,2,...,6$ if$\exists i: S'_i \cap T_i\ne \phi$ done so let $\forall i: S'_i \cup T_i=\phi$$\implies$$\sum_{i=1}^6 |S'_i \cup T_i|=$ $\sum_{i=1}^6 |S'_i|+|T_i|-|S'_i \cap T_i|$=$\sum_{i=1}^6 |S'_i|+|T_i|$ $=\sum_{i=1}^6 |S'_i| + \sum_{i=1}^6 |T_i|=$ $1978-6 + 1978-6= 1972+1972$ But we have only $1978$ numbers. $\implies$$\exists i: S'_i \cap T_i\ne \phi$ done

Is it true?

My solution: Suppose the opposite. We have $1978=6*329+4$, thus there is at least one country containing at least $330$ members (call it $A$). Let $330$ members of $A$ be $a_{1}<a_{2}<...<a_{330}$ ($i-th$ member is labeled with $a_i$). Consider people labeled with $a_2-a_1,...,a_{330}-a_1$. There are $329$ of them and none of them belong to $A$ (If so, then we have $a_i-a_1=a_j$ thus $a_j=a_i+a_1$, which is contradiction with my assumption). But $329=65*5+4$, thus there is country $B$ containing at least $66$ members. Let those members we labeled with $b_{1}<...<b_{66}$. Now consider numbers $b_2-b_1,...,b_{66}-b_1$. None of them belong to $A$ or $B$ thus they are contained in rest of $4$ countries. Also $66=4*16+2$ thus there is country $C$ containing at least $17$ members say $c_1,...,c_{17}$. Consider numbers $c_2-c_1,...,c_{17}-c_1$ (they don't belong to $A,B$ or $C$). There are $16$ of them and $16=3*5+1$ thus there is at least $6$ members in country $D$, say $d_1<...<d_6$. Consider $d_2-d_1,...,d_6-d_1$. There are $5$ of them, contained in rest of $2$ countries. There's a country containing at least $2$ of them, say numbers $e_1<e_2$. Then consider number $e_2-e_1$. It belongs nowhere, but we have $e_2-e_1 \in \{1,2,3,...1978 \}$. Thus, contradiction, so there is at least one member whose number is the sum of the numbers of two members from his own country, or twice as large as the number of one member from his own country. Q.E.D