Question

How many ordered four-tuples of integers (a, b, c, d) with 0 < a < b < c < d < 100 satisfy a + d = b + c and bc − ad = 93?​

Answer 1

Step-by-step explanation:

Let $k = a + d = b + c$ so $d = k-a, b=k-c$. It follows that $(k-c)c - a(k-a) = (a-c)(a+c-k) = (c-a)(d-c) = 93$. Hence $(c - a,d - c) = (1,93),(3,31),(31,3),(93,1)$.

Solve them in terms of $c$ to get $(a,b,c,d) = (c - 93,c - 92,c,c + 1),$ $(c - 31,c - 28,c,c + 3),$ $(c - 1,c + 92,c,c + 93),$ $(c - 3,c + 28,c,c + 31)$. The last two solutions don't follow $a < b < c < d$, so we only need to consider the first two solutions.

The first solution gives us $c - 93\geq 1$ and $c + 1\leq 499$ $\implies 94\leq c\leq 498$, and the second one gives us $32\leq c\leq 496$.

So the total number of such quadruples is $405 + 465 = \boxed{870}$.