2025 SSMO Team Round Problems/Problem 9

Problem

Pairwise distinct integers $a,$ $b,$ $c,$ and $d$ satisfy the system of equations $\begin{align*} ab &= cd \\ a+d &= b+c \\ \tfrac{1}{a} + \tfrac{1}{b} - \tfrac{1}{5} &= \tfrac{1}{c} + \tfrac{1}{d} + \tfrac{1}{5}. \end{align*}$ What is the minimum possible value of $a^2+b^2+c^2+d^2$ ?

Solution

Rearrange the second equation as $a-b = c-d$ and square it to obtain $a^2-2ab+b^2=c^2-2cd+d^2$ . Adding $4ab=4cd$ to this equation, we obtain $(a+b)^2 = (c+d)^2$ , or $a+b = \pm (c+d)$ . For the sake of contradiction, suppose $a+b = c+d$ . Because $ab=cd$ , by Vieta, we must have that $\{a,b\}=\{c,d\}$ , which violates the condition that the four variables be pairwise distinct. Thus, we must have $a+b = -c-d$ . This implies that $a+d = -b-c = b+c$ , so $b+c=a+d=0$ . Hence, $(a,b,c,d)$ is of the form $(u,v,-v,-u)$ , where $u$ and $v$ are distinct nonzero integers.

Any quadruple $(a,b,c,d)= (u,v,-v,-u)$ satisfies the first two equations in the problem statement, so it suffcies to determine the minimum possible value of $2(u^2+v^2) = a^2+b^2+c^2+d^2$ over the ordered pairs $(u,v)$ satisfying the equation $\frac{1}{u}+\frac{1}{v}-\frac{1}{5} = -\frac{1}{u}-\frac{1}{v}+\frac{1}{5}.$ Clearing the fractions and using SFFT, we obtain $(u-5)(v-5) = 25$ . Keeping in mind that $u,v\ne 0$ and $u\ne v$ , the only solutions to this equation in the integers are $(u,v) = (30,6),(-20,4)$ and permutations. By inspection, the minimum possible value of $2(u^2+v^2)$ is $2((-20)^2+4^2) = \boxed{832}$ .

~Sedro

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2025 SSMO Team Round Problems/Problem 9

Problem

Solution