Lovable series

$a_{n+1} = 100 + (n + 1)^2 = 100 + n^2 + 2n + 1 = a_{n} + (2n + 1)$

Now it is asked to find $\gcd(a_n, a_{n+1}) = \gcd((2n + 1), a_n) = \gcd(2n + 1, 100 + n^2)$ .

Now we need to see if $(2n + 1)$ divides $100 + n^2$

So by long division method:-

$100 + n^2 = (2n + 1)\left(\dfrac{2n - 1}{4}\right) + \dfrac{401}{4}$

$\therefore \dfrac{n^2 + 100}{2n + 1} = \left(\dfrac{2n - 1}{4}\right) + \dfrac{401}{4(2n + 1)}$

$= \dfrac{1}{4} \left((2n - 1) + \dfrac{401}{2n + 1}\right)$

Note that $401$ is a prime number so its factors are $1$ and $401$

so $2n + 1 = 1$ or $2n + 1 = 401$

$\therefore n = 0, 200$

When substituting $n = 0$ we get $100$ which is a integer.

Similarly substituting $n = 200$ we get $100$

Which means $2n + 1$ divides $100 + n^2$ when $n = 0, 200$

Therefore $2n + 1 = 1, 401$ at $n = 0, 200$

$\therefore max(d_n) = \gcd(a_n, a_{n+1}) = 401$