Maths Olympiad Prep

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, 2012

Algebra Difficulty 5.9 AIME, harder Prove it Belarus

Non-zero real numbers aa, bb, cc, dd satisfy the equalities
a+b+c+d=0,1a+1b+1c+1d+1abcd=0. a + b + c + d = 0, \quad \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{1}{abcd} = 0.
Find all possible values of the product (ab−cd)(c+d)(ab - cd)(c + d).

Solution

Answer: −1-1.
By condition,
1a+1b+1c+1d=−1abcd⇒bcd+cda+dab+abc=−1. \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = -\frac{1}{abcd} \Rightarrow bcd + cda + dab + abc = -1.
So
−1=bcd+cda+dab+abc=(bcd+cda)+(dab+abc)=cd(b+a)+ab(c+d)==[a+b+c+d=0⇒a+b=−(c+d)]=ab(c+d)−cd(c+d)=(ab−cd)(c+d). \begin{aligned} -1 = bcd + cda + dab + abc &= (bcd + cda) + (dab + abc) = cd(b + a) + ab(c + d) = \\ &= [a + b + c + d = 0 \Rightarrow a + b = -(c + d)] = ab(c + d) - cd(c + d) = (ab - cd)(c + d). \end{aligned}
Therefore, (ab−cd)(c+d)=−1(ab - cd)(c + d) = -1 for all admissible values of aa, bb, cc, dd.

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