That's a lot of digits

Note that powers of $n$ are cyclical $\mod n^2+1$ :

$(n \equiv n) \mod {n^2+1}$

$(n^2 \equiv -1) \mod {n^2+1}$

$(n^3 \equiv -n) \mod {n^2+1}$

$(n^4 \equiv n) \mod {n^2+1}$

Therefore, under $\mod n^2+1$ , we can divide the numbers into groups of $4$ powers of $n$ . Rewriting the number:

$b +na+n^2b+n^3a+n^4b+...+n^{2c-1}b+n^{2c}a$

Looking at only the first group of $4$ powers of $n$ , and taking $\mod n^2+1$ :

$b +na+n^2b+n^3a \mod n^2+1$

Using the equivalencies outlined in the beginning:

$b +na-b-na \mod n^2+1$

$0 \mod n^2+1$

Each group of $4$ powers of $n$ will therefore cancel itself out under $\mod n^2+1$ . Since there are an even amount of $ab$ groups, the number of digits will be a multiple of $4$ . This means that the number will be divided into groups of $4$ powers of $n$ , without anything left over. Since these groups are all $0 \mod n^2+1$ , the number will be $0 \mod n^2+1$ . The number is divisible: $\boxed{True}$ .