Find the remainder from the division of $3^{2017}-1$ into $3^{403}-1$

Question

Here is an interesting problem:

Find the remainder from the division of $3^{2017}-1$ into $3^{403}-1$

Answer 1

Hint :

$$3^{2017} = \Big(3^{403}\Big)^5 \cdot 3^2$$

Answer 2

$$3^{403}\equiv 1\pmod{3^{403}-1}$$ Raise to the power 5. You get,

$$3^{2015}\equiv 1\pmod{3^{403}-1}$$

$$3^{2017}\equiv 9\pmod{3^{403}-1}$$

$$3^{2017}-1\equiv 8\pmod{3^{403}-1}$$

So the remainder is 8.

Answer 3

If $n=3^{403}$, you are dividing $9n^5-1$ by $n-1$.

But $9n^5-1=9(n^5-1)+8$ and the remainder is $8$.