Using Fermat's Little Theorem

Number Theory Level 3

$\large (n+2017)^{p}-n^{p}-2016^{p}-1$

Assuming that $p$ is a prime number and $n$ is an arbitrary natural number, find the remainder when the expression above is divided by $p$ .

Cannot be determined

p-1

1

2017

0

n-1

2016

1 solution

Tommy Li
Jul 22, 2016

$\large (n+2017)^{p}-n^{p}-2016^{p}-1 \pmod{p}$

$\equiv n+2017-n-2016-1 \pmod{p}$

$\equiv 0 \pmod{p}$

What about p being 2 ?

Aniruddha Bagchi - 4 years, 4 months ago

It is of no problem

Prithwish Mukherjee - 2 years, 5 months ago