Question

[ĐTT]

Cho các cặp số nguyên \(x;y;z;t\) thỏa mãn \(\dfrac{x+y}{y+z}=\dfrac{y+z}{z+t}=\dfrac{z+t}{t+x}=\dfrac{t+x}{x+y}\)

Chứng tỏ rằng biểu thức \(A=\left(\dfrac{y+z}{x+t}\right)^{2013}+\left(\dfrac{y+t}{x+y}\right)^{2014}\) có giá trị nguyên

Answer 1

[Akai Haruma]

Lời giải:

\(\frac{x+y}{y+z}=\frac{y+z}{z+t}=\frac{z+t}{t+x}=\frac{t+x}{x+y}\)

\(\Rightarrow (\frac{x+y}{y+z})^4=(\frac{y+z}{z+t})^4=(\frac{z+t}{t+x})^4=(\frac{t+x}{x+y})^4=\frac{x+y}{y+z}.\frac{y+z}{z+t}.\frac{z+t}{t+x}.\frac{t+x}{x+y}=1\)

\(\Rightarrow \left[\begin{matrix} \frac{x+y}{y+z}=\frac{y+z}{z+t}=\frac{z+t}{t+x}=\frac{t+x}{x+y}=1\\ \frac{x+y}{y+z}=\frac{y+z}{z+t}=\frac{z+t}{t+x}=\frac{t+x}{x+y}=-1\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} x=y=z=t\\ x+y+z+t=0\end{matrix}\right.\)

Nếu $x=y=z=t$ thì:

\(A=\left(\frac{y+z}{x+t}\right)^{2013}+\left(\frac{y+t}{x+y}\right)^{2014}=\left(\frac{x+x}{x+x}\right)^{2013}+\left(\frac{x+x}{x+x}\right)^{2014}=1+1=2\in\mathbb{Z}\)

Nếu $x+y+z+t=0$ thì:

\(y+z=-(x+t); y+t=-(x+y)\)

\(\Rightarrow A=(-1)^{2013}+(-1)^{2014}=(-1)+1=0\in\mathbb{Z}\)

Vậy biểu thức $A$ luôn có giá trị nguyên.