Problem 3. Consider a natural number $n$. We say that a triplet of non-zero natural numbers, not necessarily distinct, $(x, y, z)$ is of type $n$ if $x+y+z=n$ and we denote by $s(n)$ the number of triplets of type $n$.

a) Show that there is no natural number $n$ for which $s(n)=14$.

b) Determine the smallest natural number $n$ for which $s(n)>2014$.
Solution. a) Three different numbers generate 6 triplets. If only two of the numbers in the triplet are equal, 3 such triplets can be formed. A triplet with all components equal can be formed at most once. It follows that the number of triplets is of the form $3k$ or $3k+1, k \in \mathbb{N}$, and 14 is of the form $3k+2$. In conclusion, there are no natural numbers that satisfy the condition $3p$.

b) For $x=1$, we get $y+z=n-1$ and there are $n-2$ triplets.

For $x=2$, we get $y+z=n-2$ and there are $n-3$ triplets.

For $x=n-2$, we get $y+z=2$ and there is 1 triplet.

The total number of triplets is $1+2+\ldots+(n-2)=\frac{(n-2)(n-1)}{2}$

From the condition $(n-2)(n-1) \geq 4030$ and $n$ being the smallest natural number, we get $n=65$.