To satisfy an equation

Consider the equation modulo $3$ ; we get

$n-n \equiv 1+1$

This clearly has no solutions.

${2017}^{2017} + 1 = (3 \times 672 +1)^{2017} + 1 \equiv 1^{2017} + 1 \equiv 2 \mod 3 \\ \therefore 3 \nmid {2017}^{2017} + 1$

• if $n = 3k$

$n^3 + 2015n = n(n^2 + 2015) = 3 \times k(n^2 + 2015) \\ \therefore 3 | n^3 + 2015n$

• if $n = 3k + 1$

$n^3 + 2015n = n((3k+1)^2 + (3 \times 671 +2)) = n(9k^2 + 6k + 1 + 3 \times 671+ 2) = 3 \times n(3k^2 + 2k + 672) \\ \therefore 3 | n^3 + 2015n$

• if $n = 3k + 2$

$n^3 + 2015n = n((3k+2)^2 + (3 \times 671 +2)) = n(9k^2 + 12k + 4 + 3 \times 671+ 2) = 3 \times n(3k^2 + 4k + 673) \\ \therefore 3 | n^3 + 2015n$

So $3 | n^3 + 2015n$ , then no solution.