Question

Cho x,y,z là các số tự nhiên thỏa mãn điều kiện \(\frac{x-1}{2}=\frac{y+5}{5}=\frac{z-2}{3}\) và \(3x+2y-5z+16=0\). Chứng minh rằng: \(P=x^{2019}+y^{2019}+z^{2019}\) chia hết cho 4.

Answer 1

\(\frac{3x-3}{6}=\frac{2y+10}{10}=\frac{5z-10}{15}=\frac{3x+2y-5z+17}{1}=\frac{3x+2y-5z+16+1}{1}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x-1}{2}=1\\\frac{y+5}{5}=1\\\frac{z-2}{3}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=0\\z=5\end{matrix}\right.\)

\(\Rightarrow P=3^{2019}+5^{2019}\)

Ta có \(3\equiv-1\left(mod4\right)\Rightarrow3^{2019}\equiv-1\left(mod4\right)\)

\(5\equiv1\left(mod4\right)\Rightarrow5^{2019}\equiv1\left(mod4\right)\)

\(\Rightarrow P\equiv\left(-1+1\right)\left(mod4\right)\Rightarrow P\equiv0\left(mod4\right)\Rightarrow P⋮4\)