$\{1, 2, . . . , 49\}$ is partitioned into $3$ subsets. Prove that at least $1$ of them contains $3$ different numbers $a,b,c$ such that $a+b=c$ As the title says, I'm asked to prove that at least one of the subsets contains three different numbers $a, b$ and c such that $a + b = c$. I tried to prove it but it is more difficult than it initially seemed to be. How would you do it?
Edit: The original set is arbitrarily partitioned.
Edit2: The exercise was given in a first year Discrete Mathematics course. I'm pretty sure they want us to somehow use the pigeonhole principle.
 A: An informal proof with contradiction:
Assume that there is no such subset.
By pigeonhole principle, there must be one subset (of the three), say $A$, has at least $17$ elements, say $a_1<a_2<...<a_{17}$. Form assumption we know if the difference $a_{17}-a_i \neq a_i$, $i=1,2,...,16$, then it will not belong to $A$. There can be at most one $i$ s.t. $a_{17}-a_i=a_i$.
Then, by pigeonhole principle again, one of the rest two subsets, say $B$, must have at least $8$ of those differences $a_{17}-a_i$ which are not in $A$, say $b_1<b_2<...<b_8$, and similarly, if the difference $b_8-b_j \neq b_j$, $j=1,2,...,7$, then it will not belong to $B$. There can be at most one $j$ s.t. $b_8-b_j=b_j$.
For any $j$ there won't be such $j' \neq j$ that $b_{j'}-b_j \in A$. If there was one, then $b_j=a_{17}-a_{i'}$ and $b_j=a_{17}-a_i$, then $(a_{17}-a_{i'})-(a_{17}-a_i)=a_i-a_{i'} \in A$ where $i \neq i'$, a contradiction.
So those $8$ (or $7$) differences must be in the last subset, say $C$. We can order them and denote them by $c_1<c_2<...<c_7$ (or $c_8$). But similarly, only at most one of their differences $c_8-c_k$ (or $c_7-c_k$) will lie in $C$. The others are not in $B$, either. While If $c_8-c_k \in A$ then there is $b_j-b_{j'} \in A$, and this is impossible, thus, they are not in $A$.
This contradicts with the fact that $A$, $B$ and $C$ consist of a partition. 
