For the first part, if p = q then teh equations are identical so either a = c and b = d, or a= d and b = c. In either case there is a factor in the expression that is zero (either a - d or b - d). So (a + c)(b + c)(a – d)(b – d) = 0 if p = q.

For the second part, note that the information in the question tells us that the first equation can be rewritten as

(x – a)(x – b) = 0 , or x² – (a + b)x + ab = 0.

Hence, a + b = - p and ab = 1. Similarly, c + d = - q and cd = 1.

Secondly, note that by substituting c and d in the second equation, we see that

qc = – (c² + 1), and qd = – (d² + 1).

Multiplying out the four brackets in the problem gives us

(a + c)(b + c)(a – d)(b – d) =

(ab + bc + ac + c²)(ab – ad – bd + d²) =

(1 – pc + c²)(1 + pd + d²) =

( – qc – pc )( – qd + pd) =

c (q + p) d (q – p) =

cd(q² – p²) = q² – p²

Note how beneficial it is to use the conditions worked out at the start to simplify the algebra as you go along!

This answer also makes perfect sense because it comes to zero when p = q, because in that case, the solutions a and b must be the same as the solutions c and d, so at least one of a – d and b – d must be zero.

Challenge 264: Quadratic Query