Problem (Number Theory):
An integer-valued function $f$ is called tenuous if $f(x)+f(y)>x^{2}$ for all positive integers $x$ and $y$. Let $g$ be a tenuous function such that $g(1)+g(2)+\cdots+g(20)$ is as small as possible. Compute the minimum possible value for $g(14)$.

Solution:
For a tenuous function $g$, let $S_{g}=g(1)+g(2)+\cdots+g(20)$. Then:

$$
\begin{aligned}
S_{g} & =(g(1)+g(20))+(g(2)+g(19))+\cdots+(g(10)+g(11)) \\
& \geq\left(20^{2}+1\right)+\left(19^{2}+1\right)+\cdots+\left(11^{2}+1\right) \\
& =10+\sum_{k=11}^{20} k^{2} \\
& =2495 .
\end{aligned}
$$



The following argument shows that if a tenuous function $g$ attains this sum, then $g(1)=$ $g(2)=\cdots=g(10)$. First, if the sum equals 2495 , then $g(1)+g(20)=20^{2}+1, g(2)+g(19)=$ $19^{2}+1, \ldots, g(10)+g(11)=11^{2}+1$. If $g(1)<g(2)$, then $g(1)+g(19)<19^{2}+1$, which contradicts the tenuousness of $g$. Similarly, if $g(2)>g(1)$, then $g(2)+g(20)<20^{2}+1$. Therefore $g(1)=g(2)$. Analogously, comparing $g(1)$ and $g(3), g(1)$ and $g(4)$, etc. shows that $g(1)=g(2)=g(3)=\cdots=g(10)$.

Now consider all functions $g$ for which $g(1)=g(2)=\cdots=g(10)=a$ for some integer $a$. Then $g(n)=n^{2}+1-a$ for $n \geq 11$. Because $g(11)+g(11)>11^{2}=121$, it is the case that $g(11) \geq 61$. Thus $11^{2}+1-a \geq 61 \Rightarrow a \leq 61$. Thus the smallest possible value for $g(14)$ is $14^{2}+1-61=\mathbf{1 3 6}$.

Final answer: 136