One-rooted Polynomials

The equation rearranges to $x^3 + 3x + 5 - (x - c)^3 = (ax + b)^2$ Say the LHS is $rx^2 + sx + t$ . We need this to be a perfect square. Therefore, we must have $s^2 - 4st = 0$ , i.e. $s^2 = 4st$ . Calculating the values of $r, s, t$ in terms of $c$ , we get $(3 - 3c^2)^2 = 4(-3c)(5 - c^3)$ which simplifies to $c^4 + 6c^2 - 20c - 3 = 0 \tag{1}$ Now suppose we have any value of $c$ satisfying this quartic equation. Then we will have remaining to choose $a$ and $b$ such that $rx^2 + sx + t = (ax + b)^2$ , given that $s^2 = 4st$ . Observe that $r \ne 0$ , because if it were we would have $-3c = 0$ , or $c = 0$ , which does not satisfy (1). Therefore, $rx^2 + sx + t = (ax + b)^2$ has exactly two solutions for $a$ and $b$ -- first, there are two values of $a$ such that $a^2 = r$ , and then once $a$ is chosen, $b$ is determined.

In summary, for each value of $c$ satisfying (1), there are exactly two possible pairs $(a,b)$ . You can check that (1) does not have any double roots by showing that its derivative and itself share no roots in common. Hence (by factoring over the complex numbers), there are 4 valid values of $c$ , for a total of $4 \cdot 2= \boxed{8}$ triples $(a,b,c)$ of complex numbers.