Answer
128
Work Step by Step
Let the numbers be $x$ and $y$, so $x+y=16$, and $y=16-x$. Call the sum of the squares $S$, so $S=x^{2}+(16-x)^{2}$.
$S'=4x-32$, so $8$ is a critical value and minimizes the function. Therefore the minimum value of the squares is $8^{2}+(16-8)^{2}=64+64=128.$
