April wrote: Consider a subset $ A$ of 84 elements of the set $ \{1,\,2,\,\dots,\,169\}$ such that no two elements in the set add up to 169. Show that $ A$ contains a perfect square. I think it is obvious at all from $ 169=5^2+12^2$ and $ 169=2.84+1$ If the set doesn't contain 25 or 144 it will contain a pairs of $ (k,169-k)$ ,it gives contradiction .

In other words, consider the family of $84$ doubletons $D_k = \{k, 169-k\}$, for $1\leq k \leq 84$. If $169 \in A$, then we are done, since $169 = 13^2$. If not, since then $A \subset \{1,2,\ldots,168\} = \bigcup_{k=1}^{84} D_k$, and we need have $|A \cap D_k| \leq 1$ for all $1\leq k \leq 84$, it follows $|A \cap D_k| = 1$ for all $1\leq k \leq 84$. But $D_{25} = \{25, 144\} = \{5^2,12^2\}$, and so $A$ will contain a perfect square.