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Ta có 

(m+n+p)^q >= m^q+n^q+p^q

=>a+b+c=1

=>(a+b+c)^2016=1 >= a²⁰¹⁶ + b²⁰¹⁶ + c²⁰¹⁶

^Mà  a²⁰¹⁶ + b²⁰¹⁶ + c²⁰¹⁶ >=0

=>  a²⁰¹⁶ + b²⁰¹⁶ + c²⁰¹⁶=1

\(a+b=x+y\)

\(\Rightarrow a-x=y-b\) (1)

\(a^2+b^2=x^2+y^2\)

\(\Rightarrow a^2-x^2=y^2-b^2\)

\(\Leftrightarrow\left(a-x\right)\left(a+x\right)=\left(y-b\right)\left(y+b\right)\)

\(\Leftrightarrow\left(a-x\right)\left(a+x\right)-\left(a-x\right)\left(y+b\right)=0\)

\(\Leftrightarrow\left(a-x\right)\left[\left(a+x\right)-\left(y+b\right)\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-x=0\\\left(a+x\right)-\left(y+b\right)=0\end{matrix}\right.\)

Với \(a-x=0\) , kết hợp với (1) ta được:

\(\left\{{}\begin{matrix}a-x=y-b\\a-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b=y\\a=x\end{matrix}\right.\)

\(\Rightarrow a^{2016}+b^{2016}=x^{2016}+y^{2016}\)

Với \(a-x=y-b\)

\(\left\{{}\begin{matrix}a+b=x+y\\a+x=y+b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=y\\b=x\end{matrix}\right.\)

\(\Rightarrow a^{2016}+b^{2016}=x^{2016}+y^{2016}\)

$\textbf{1.}$

Ta có $5^{2017}+5^{2015}=5^{2015}(5^2+1)$

$=5^{2015}\cdot26$

$=5^{2015}\cdot13\cdot2.$

Vì $13\mid13\cdot2$ nên $13\mid\left(5^{2017}+5^{2015}\right).$

Vậy $5^{2017}+5^{2015}$ chia hết cho $13.$

$\textbf{2.}$

Giả sử $a^{2014}+b^{2015}+c^{2016}\vdots6.$

Ta có $a^{2016}-a^{2014}=a^{2014}(a^2-1)$

$=a^{2014}(a-1)(a+1).$

Vì $a(a-1)(a+1)\vdots6$ nên $a^{2014}(a-1)(a+1)=a^{2013}\cdot a(a-1)(a+1)\vdots6.$

Suy ra $a^{2016}\equiv a^{2014}\pmod6.$

Tương tự, $b^{2017}-b^{2015}=b^{2015}(b^2-1)$

$=b^{2014}\cdot b(b-1)(b+1)\vdots6,$ nên $b^{2017}\equiv b^{2015}\pmod6.$

Lại có $c^{2018}-c^{2016}=c^{2016}(c^2-1)$

$=c^{2015}\cdot c(c-1)(c+1)\vdots6,$ nên $c^{2018}\equiv c^{2016}\pmod6.$

Cộng ba đồng dư trên, $a^{2016}+b^{2017}+c^{2018}\equiva^{2014}+b^{2015}+c^{2016}\equiv0\pmod6.$

Vậy $a^{2016}+b^{2017}+c^{2018}$ chia hết cho $6.$