CMIMC contest question (2)

A substitution yields $x=(x^2+15x+32)^2+49(x^2+15x+32)+593\Rightarrow x^4+30x^3+338x^2+1694x+3185=0$ . Since $3185=5\cdot 7^2\cdot13$ , there are only $24$ possible rational roots. Testing them all, one finds $-7$ is a root (it is also easily checked that $-7$ is a double root and that the other two roots are imaginary). Therefore, the parabolas intersect at the point $(-7,-24)$ .

Since we know the two parabolas intersect, we can write the given equations as $y= (y^{2}+49y+593)^{2} + 15(y^{2}+49y+593) + 32$ , $\implies$ then the two equations after equating them can be simplified as:

$-58848y-y^{4}-98y^{3}-3602y^{2}-360576= 0$ . We notice that $y=-24$ is a zero of this equation, the other roots are complex numbers.

Taking $y$ as $y=-24$ , then by plugging it in $y=x^{2}+15x+32= -24$ and solving for $x$ , gives $x=-7$ , thus, $x_oy_o=168$