24. Let $a, b, c, d$ be 4 distinct nonzero integers such that $a+b+c+d=0$ and the number $M=(b c-a d)(a c-b d)(a b-c d)$ lies strictly between 96100 and 98000 . Determine the value of $M$.
24. Answer: 97344
Solution. First we show that $M$ must be a square. Let $d=-a-b-c$. Then $b c-a d=$ $b c+a(a+b+c)=(a+c)(a+b), a c-b d=a c+b(a+b+c)=(b+c)(b+a)$, and $a b-c d=a b+c(a+b+c)=(c+a)(c+b)$. Therefore $M=(a+b)^{2}(b+c)^{2}(c+a)^{2}$. Note that $(a+b)(b+c)(c+a)$ cannot be an odd integer since two of the 3 numbers $a, b, c$ must be of the same parity. The only squares in $(96100,98000)$ are $311^{2}, 312^{2}, 313^{2}$. Since 311 and 313 are odd, the only value of $M$ is $312^{2}=97344$. When $a=18, b=-5, c=6, d=-19$, it gives $M=97344$.