Solve in integers the system of equations $$\begin{array}{lll} x^2y+y^2z+z^2x-3xyz&=&23 \\ xy^2+yz^2+zx^2-3xyz&=&25. \end{array}$$
Proposed by Adrian Andreescu, Dallas, USA
Solution:
Subtracting the first equation to the second equation, we have
$$xy(y-x)+yz(z-y)+zx(x-z)=2,$$ i.e. $$(x-y)(y-z)(z-x)=2.$$ Since $(x-y)+(y-z)+(z-x)=0$ and $2+(-1)+(-1)=(-2)+1+1=0$, then
$$\begin{array}{lll} x-y&=&-1 \\ y-z&=&-1 \\ z-x&=&2, \end{array} \qquad \begin{array}{lll} x-y&=&-1 \\ y-z&=&2 \\ z-x&=&-1, \end{array} \qquad \begin{array}{lll} x-y&=&2 \\ y-z&=&-1 \\ z-x&=&-1, \end{array}$$
$$\begin{array}{lll} x-y&=&1 \\ y-z&=&1 \\ z-x&=&-2, \end{array} \qquad \begin{array}{lll} x-y&=&1 \\ y-z&=&-2 \\ z-x&=&1, \end{array} \qquad \begin{array}{lll} x-y&=&-2 \\ y-z&=&1 \\ z-x&=&1. \end{array}$$
Hence, $$\begin{array}{lll} (x,y,z) & \in & \{(n,n+1,n+2),(n,n+1,n-1),(n,n-2,n-1),\\& & (n,n-1,n-2),(n,n-1,n+1),(n,n+2,n+1)\}. \end{array}$$
Substituting each triple into the first equation of the system and with an easy check to the second equation, we get
$$(x,y,z) \in \{(7,8,9),(8,9,7),(9,7,8)\}.$$
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