A Decade of Divisibility

We shall prove that, if positive integers $m, n$ are relatively prime, then

$mn|m^{\phi(n)} + n^{\phi(m)} - 1,$

where $\phi$ represents Euler's totient function.

Taking the expression modulo $m$ and $n$ individually and applying Euler's Totient Theorem, we find that

$m^{\phi(n)} + n^{\phi(m)} - 1 \equiv n^{\phi(m)} - 1 \equiv 1 - 1 \equiv 0 \! \! \! \! \pmod{m}$

and

$m^{\phi(n)} + n^{\phi(m)} - 1 \equiv m^{\phi(n)} - 1 \equiv 1 - 1 \equiv 0 \! \! \! \! \pmod{n}.$

Since $m$ and $n$ are relatively prime, we can apply the Chinese Remainder Theorem to conclude that $m^{\phi(n)} + n^{\phi(m)} - 1 \equiv 0 \! \pmod{mn},$ or $mn|m^{\phi(n)} + n^{\phi(m)} - 1. \, \, \blacksquare$

Now, we can apply this result to the problem. As $2017$ and $2027$ are prime, they are also relatively prime. Also, $\phi(2017) = 2016$ and $\phi(2027) = 2026.$ Thus,

$2017^{2026} + 2027^{2016} - 1 \equiv \boxed{0} \! \! \! \! \pmod{N}.$