Question

Giải phương trình nghiệm nguyên: \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=y^2\)

Answer 1

Ta có: \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=y^2\)

<=> \(\left(x^2+3x\right)\left(x^2+3x+2\right)=y^2\)

Đặt \(x^2+3x=a\) (a \(\in\) Z)

=> \(x^2+3x+2=a+2\) (*)

Thay (*) vào PT ta được:

\(a\left(a+2\right)=y^2\)

<=> \(a^2+2a=y^2\)

<=> \(\left(a^2+2a+1\right)-1=y^2\)

<=> \(\left(a+1\right)^2-y^2=1\)

<=> \(\left(a+1-y\right)\left(a+1+y\right)=1\)

Xét 2 TH

+ TH₁: \(\left\{{}\begin{matrix}a+1-y=1\\a+1+y=1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}a-y=0\\a+y=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=0\\y=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x^2+3x=0\\y=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=-3;0\\y=0\end{matrix}\right.\) (TM)

+ TH₂: \(\left\{{}\begin{matrix}a+1-y=-1\\a+1+y=-1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a+y=-2\\a-y=-2\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a=-2\\y=0\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}x^2+3x+2=0\\y=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\left(x+1\right)\left(x+2\right)=0\\y=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=-1;-2\\y=0\end{matrix}\right.\) (TM)

Vậy nghiệm nguyên (x;y) = (-1;0), (-2;0), (0;0), (-3;0)

Answer 2

Ở đây nè bạn:https://hoc24.vn/hoi-dap/question/278218.html