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− | == Solution== | + | ==Solution== |
− | 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
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− | <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>).
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| == See also == | | == See also == |