Giải phương trình trên tập số nguyên 

- Hướng dẫn giải

$x^{2015}=\sqrt{y(y+1)(y+2)(y+3)}+1$ (1)

$$\begin{gathered} y(y+1)(y+2)(y+3)=\left[y(y+3)\right]\left[(y+1)(y+2)\right]=(y^2+3y)(y^2+3y+2) \\ Đặt t=y^2+3y+1\Rightarrow y(y+1)(y+2)(y+3)=t^2-1 \end{gathered}$$

( t ∈ ℤ , t² ≥ 1)

$\left(1\right) \iff x^{2015}-1=\sqrt{t^2-1}\iff \left\{\begin{array}{l}{x}^{2015}-1\ge 0\\ \left(x^{2015}-1\right)=t^2-1\left(2\right)\end{array}\right.$

Với x, t là số nguyên ta có: 

$\begin{array}{l}\left(2\right)\iff \left(x^{2015}-1+t\right)\left(x^{2015}-1-t\right)=-1\\ \iff \left[\begin{array}{l}\left\{\begin{array}{l}{x}^{2015}-1+t=1\\ x^{2015}-1-t=-1\end{array}\right.\\ \left\{\begin{array}{l}{x}^{2015}-1+t=-1\\ x^{2015}-1-t=1\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}{x}^{2015}=t=1\\ \left\{\begin{array}{l}{x}^{2015}=1\\ t=-1\end{array}\right.\end{array}\right.\end{array}$

Với $x^{2015}=t=1\Rightarrow \left\{\begin{array}{l}x=1\\ y^2+3y+1=1\end{array}\right.\iff \left\{\begin{array}{l}x=1\\ \left[\begin{array}{l}y=0\\ y=-3\end{array}\right.\end{array}\right.$

Với $\left\{\begin{array}{l}{x}^{2015}=1\\ t=-1\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=1\\ y^2+3y+1=-1\end{array}\right.\iff \left\{\begin{array}{l}x=1\\ \left[\begin{array}{l}y=-1\\ y=-2\end{array}\right.\end{array}\right.$

Thử lại ta thấy các cặp (1;-3), (1;-2), (1;-1), (1;0) thỏa mãn đề bài

Vậy có 4 cặp (x;y) cần tìm là (1;-3), (1;-2), (1;-1), (1;0)