The numbers in the [sequence](https://artofproblemsolving.com/wiki/index.php/Sequence) $101$, $104$, $109$, $116$,$\ldots$ are of the form $a_n=100+n^2$, where $n=1,2,3,\ldots$ For each $n$, let $d_n$ be the greatest common divisor of $a_n$ and $a_{n+1}$. Find the maximum value of $d_n$ as $n$ ranges through the [positive integers](https://artofproblemsolving.com/wiki/index.php/Positive_integer).
If $(x,y)$ denotes the [greatest common divisor](https://artofproblemsolving.com/wiki/index.php/Greatest_common_divisor) of $x$ and $y$, then we have $d_n=(a_n,a_{n+1})=(100+n^2,100+n^2+2n+1)$. Now assuming that $d_n$ [ divides](https://artofproblemsolving.com/wiki/index.php/Divisor) $100+n^2$, it must divide $2n+1$ if it is going to divide the entire [expression](https://artofproblemsolving.com/wiki/index.php/Expression) $100+n^2+2n+1$.
Thus the [equation](https://artofproblemsolving.com/wiki/index.php/Equation) turns into $d_n=(100+n^2,2n+1)$. Now note that since $2n+1$ is [ odd](https://artofproblemsolving.com/wiki/index.php/Odd_integer) for [ integral](https://artofproblemsolving.com/wiki/index.php/Integer) $n$, we can multiply the left integer, $100+n^2$, by a power of two without affecting the greatest common divisor. Since the $n^2$ term is quite restrictive, let's multiply by $4$ so that we can get a $(2n+1)^2$ in there.
So $d_n=(4n^2+400,2n+1)=((2n+1)^2-4n+399,2n+1)=(-4n+399,2n+1)$. It simplified the way we wanted it to!
Now using similar techniques we can write $d_n=(-2(2n+1)+401,2n+1)=(401,2n+1)$.  Thus $d_n$ must divide $\boxed{401}$ for every single $n$.  This means the largest possible value for $d_n$ is $401$, and we see that it can be achieved when $n = 200$.