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Difference between revisions of "2018 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 9"

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

Revision as of 02:20, 26 January 2022

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 $(4, 4, 4, 4)$).

See also

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
All UNM-PNM Problems and Solutions
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