Question

The number of solutions $(x,y,z)$ to the system of equations $x+2y+4z=9,4yz+2xz+xy=13,xyz=3$ such that at least two of $x,y,z$ are integers is

• A. $3$
• B. $5$
• C. $6$
• D. $4$

Answer 1

Answer: (B)

$5$

Let the roots of the system equation are:

$\alpha =x,\beta =2y,\gamma =4z$

$\alpha +\beta +\gamma =x+2y+4z=9$

$\alpha \beta +\beta \gamma +\gamma \alpha =2xy+8yz+yzx$

$=2(4yz+2xz+xy)\Rightarrow 26$

$\alpha \beta \gamma =8xyz\Rightarrow 24$

Thus,our polynomial should be:

$P^3-9P+26P-24=0$

$(P-2)(P-3)(P-4)=0$

since our roots are :

$\alpha =x,\beta =2y$ and $\gamma =4z$

$(x,2y,4z)=(2,3,4)$ or its permutations,or 6 combination.

However,note that one case if,

$x=4,2y=3$ and $4z=2$

$(x,y,z)=(4,\frac{3}{2},\frac{1}{2})$

which two of the roots are not an integer :Excluding of this case ,we have five solutions.