Mystery in the Middle 2

Let the values in the vertical line of three circles be $u,y,v$ (so $y$ is the question mark).

The equations we have are $u+y=3,\;\;\;v+y=17,\;\;\;uvy=x$

We want to get an equation for $y$ in terms of $x$ that has three distinct integer roots. Substituting $u=3-y$ and $v=17-y$ : $y(3-y)(17-y)=x$

which after tidying becomes $y^3-20y^2+51y-x=0$

This is a cubic in $y$ ; say its three roots are $p,q,r$ . By Vieta, $p+q+r=20,\;\;\;qr+rp+pq=51,\;\;\;pqr=x$

Squaring the first and subtracting twice the second, we get $p^2+q^2+r^2=298$

Since $\sqrt{298}<18$ , this is easy enough to solve; the only triples (up to permutation) that work (together with $p+q+r=20$ ) are $(0,3,17)$ and $(-3,8,15)$ . The first leads to $x=0$ , which isn't allowed; hence the answer is $-3 \cdot 8 \cdot 15=\boxed{-360}$ .

If the middle number is $x$ , then the top white circle must be $(3 - x)$ and the bottom white circle must be $(17 - x)$ , and $(3 - x)x(17 - x) = X$ , which rearranges to $x^3 - 20x^2 - 51x - X = 0$ .

Suppose that there are three possible integer values $p$ , $q$ , and $r$ for the middle number $x$ . Then by Vieta's Formula , $p + q + r = 20$ , $pq + qr + pr = 51$ , and $pqr = X$ . By substitution, $p = - \frac{r}{2} + 10 \pm \frac{1}{2}\sqrt{-3r^2 + 40r + 196}$ and $q = - \frac{r}{2} + 10 \mp \frac{1}{2}\sqrt{-3r^2 + 40r + 196}$ .

For $p$ and $q$ to be integers, $-3r^2 + 40r + 196$ must be a perfect square, and for $p$ and $q$ to be real, $-3r^2 + 40r + 196 > 0$ , which solves to $-3 \leq r \leq 17$ . Testing each integer $r$ in this range, only $r = -3$ , $r = 0$ , $r = 3$ , $r = 8$ , $r = 15$ , and $r = 17$ give perfect squares for $-3r^2 + 40r + 196$ , and give the following $p = - \frac{r}{2} + 10 + \frac{1}{2}\sqrt{-3r^2 + 40r + 196}$ , $q = - \frac{r}{2} + 10 - \frac{1}{2}\sqrt{-3r^2 + 40r + 196}$ , and $X = pqr$ values:

The only non-zero $X$ -value with three possible integer values for the middle number is $X = \boxed{-360}$ .