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Let $a,b,c,d$ be $4$ distinct non-zero integers such that $a+b+c+d = 0$. It is know that the number

$$M = (bc - ad)(ac - bd)(ab-cd)$$

lies strictly between $96100$ and $98000$. Determine the value of $M$.

I tried expanding the expression out, as well as using AM-GM on it, but to no avail. Any help would be appreciated. Thanks!

(Source: Singapore Mathematical Olympiad 2013, Open Section, First Round, Question 24)

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  • $\begingroup$ If it's any help $96100=2^2\times 5^2\times 31^2$ and $98000=2^4\times5^3\times 7^2$ $\endgroup$ – Guy Mar 21 '14 at 5:05
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    $\begingroup$ After the substitution of $a = -b-c-d$, we have $M = (b+c)^2(b+d)^2(c+d)^2$. $\endgroup$ – Erik M Mar 21 '14 at 5:05
  • $\begingroup$ There are 3 perfect squares in that range, 96721, 97344, 97969 $\endgroup$ – Guy Mar 21 '14 at 5:07
  • $\begingroup$ $M=97344$ The other two are squares of primes. Erik is a genius. $\endgroup$ – Guy Mar 21 '14 at 5:09
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    $\begingroup$ @ErikMiehling post that as an answer? $\endgroup$ – Guy Mar 21 '14 at 5:09
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After the substitution of $a = -b-c-d$, we have $M = (b+c)^2(b+d)^2(c+d)^2$.

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  • $\begingroup$ upvoted. :)${}{}$ $\endgroup$ – Guy Mar 21 '14 at 5:11
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We need to say a bit more after Erik's answer. We have $$(b+c)(b+d)(c+d)=\pm311,\,\pm312,\,\pm313\ .$$ Now the sum of the three numbers $b+c$, $b+d$, $c+d$ is even and so at least one of the numbers must be even. Therefore the only possibility is that $$(b+c)(b+d)(c+d)=\pm312$$ and we have $$M=312^2\ .$$

Note: we cannot straight away rule out the possibility that the product of the three numbers is prime, because they could be $1,1,p$.

And another (possibly) interesting point. We can of course obtain Erik's identity by plain algebra, but the form of $M$ suggests determinants, so here is an alternative method. Using the fact that $a+b+c+d=0$ and adding columns, we have $$bc-ad=-\Bigl|\matrix{a&b\cr c&d\cr}\Bigr| =-\Bigl|\matrix{a+b&b\cr c+d&d\cr}\Bigr|=-\Bigl|\matrix{-(c+d)&b\cr c+d&d\cr}\Bigr|$$ Now taking out a factor of $-(c+d)$ and row-reducing, $$bc-ad=(c+d)\Bigl|\matrix{1&b\cr-1&d\cr}\Bigr| =(c+d)\Bigl|\matrix{1&b\cr0&b+d\cr}\Bigr| =(c+d)(b+d)\ ,$$ and similarly for the others.

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