, 2018

Calculating, $\left(\sqrt{4+\sqrt{4}}\right)^4 = \left(\sqrt{4+2}\right)^4 = \left(\sqrt{6}\right)^4 = \left(\left(\sqrt{6}\right)^2\right)^2 = 6^2 = 36$. Since $y$ is an integer, then $8 - y^2$ is an integer. Therefore, $\sqrt{23-x}$ is an integer which means that $23-x$ is a perfect square. Since $x$ is a positive integer, then $23 - x < 23$ and so $23-x$ must be a perfect square that is less than 23. We make a table listing the possible values of $23-x$ and the resulting values of $x$, $\sqrt{23-x} = 8-y^2$, $y^2$, and $y$: $23-x$ $x$ $\sqrt{23-x} = 8 - y^2$ $y^2$ $y$ $16$ $7$ $4$ $4$ $\pm 2$ $9$ $14$ $3$ $5$ $\pm \sqrt{5}$ $4$ $19$ $2$ $6$ $\pm \sqrt{6}$ $1$ $22$ $1$ $7$ $\pm \sqrt{7}$ $0$ $23$ $0$ $8$ $\pm \sqrt{8}$ Since $x$ and $y$ are positive integers, then we must have $(x,y) = (7,2)$. (We note that since we were told that there is only one such pair, we did not have to continue the table beyond the first row.) Since the line with equation $y=mx+2$ passes through $(1,5)$, then $5 = m + 2$ and so $m=3$. Since the parabola with equation $y = ax^2 + 5x - 2$ passes through $(1,5)$, then $5 = a + 5 - 2$ and so $a=2$. To find the coordinates of $Q$, we determine the second point of intersection of $y=3x+2$ and $y=2x^2 + 5x - 2$ by equating values of $y$: $$\begin{aligned} 2x^2 + 5x - 2 & = 3x + 2 \\ 2x^2 + 2x - 4 & = 0 \\ x^2 + x - 2 & = 0 \\ (x+2)(x-1) & = 0\end{aligned}$$ Therefore, $x=1$ or $x=-2$. Since $P$ has $x$-coordinate $1$, then $Q$ has $x$-coordinate $-2$. Since $Q$ lies on the line with equation $y=3x+2$, we have $y=3(-2)+2=-4$. In summary, (i) $m=3$, (ii) $a=2$, and (iii) the coordinates of $Q$ are $(-2,-4)$.