2018 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 9

Problem

Find the number of $4$ -tuples $(a,b,c,d)$ with $a, b, c$ and $d$ positive integers, such that $x^2-ax + b = 0, x^2-bx + c = 0, x^2 -cx + d = 0$ and $x^2-dx + a = 0$ have integer roots

Solution

Let $r_1$ , $r_2$ be the roots of the first quadratic; $r_3$ , $r_4$ the roots of the second quadratic; $r_5$ , $r_6$ the roots of the third; and $r_7$ , $r_8$ the roots of the fourth quadratic ( $r_i\in\mathbb{Z}$ for $1\leq i\leq8$ ). Using this, we can write the the four equations in terms of $r_i$ (for $1\leq i\leq8$ ): $x^2-ax+b=(x-r_1)(x-r_2)=0$ $x^2-bx+c=(x-r_3)(x-r_4)=0$ $x^2-cx+d=(x-r_5)(x-r_6)=0$ $x^2-dx+a=(x-r_7)(x-r_8)=0$ By Vieta's (or expanding out the binomials), we find that $r_1+r_2=r_7r_8=a$ $r_3+r_4=r_1r_2=b$ $r_5+r_6=r_3r_4=c$ $r_7+r_8=r_5r_6=d$ Adding all the equations together, $r_1+r_2+r_3+r_4+r_5+r_6+r_7+r_8 = r_1r_2+r_3r_4+r_5r_6+r_7r_8$ Rearrainging, $(r_1r_2-r_1-r_2)+(r_3r_4-r_3-r_4)+(r_5r_6-r_5-r_6)+(r_7r_8-r_7-r_8)=0$ Factoring with SFFT, $(r_1-1)(r_2-1)+(r_3-1)(r_4-1)+(r_5-1)(r_6-1)+(r_7-1)(r_8-1)=4$ By inspection, we find that the only solution that satisfies ALL the above equations is $r_1=r_2=r_3=r_4=r_5=r_6=r_7=r_8=2$ , which gives $\boxed{1}$ solution for $(a, b, c, d)$ (that is $(4, 4, 4, 4)$ ).

2018 UNM-PNM Contest II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
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2018 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 9

Problem

Solution

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