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Difference between revisions of "2024 SSMO Speed Round Problems/Problem 10"

(Created page with "==Problem== Let <math>a_1, a_2, \dots a_{14}</math> be the roots of <math>(x^7-x^3+2)^2=0</math>. Find the value of <math>\prod_{i=1}^{14} (a_i^7+1)</math>. ==Solution==")
 
 
(2 intermediate revisions by one other user not shown)
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Let <math>a_1, a_2, \dots a_{14}</math> be the roots of <math>(x^7-x^3+2)^2=0</math>. Find the value of <math>\prod_{i=1}^{14} (a_i^7+1)</math>.
 
Let <math>a_1, a_2, \dots a_{14}</math> be the roots of <math>(x^7-x^3+2)^2=0</math>. Find the value of <math>\prod_{i=1}^{14} (a_i^7+1)</math>.
  
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==Solution==
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==Solution 1==
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Let <math>a_1,a_2,\dots,a_7</math> be the roots of <cmath>f(x) = x^7-x^3+2 = \prod_{i=1}^7(x-a_i).</cmath> It is easy to see that <cmath>\prod_{i=1}^{14}(a_i^7+1) = \left(\prod_{i=1}^{7}(a_i^7+1)\right)^2.</cmath> Now, we have <math>a_i^7-a_i^3+2 = 0\implies a_i^7+1 = a_i^3-1.</math> Denote <math>1,\omega,\frac{1}{\omega}</math> as the roots of <math>x^3-1,</math> with <math>\omega^3 = 1.</math> So,
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<cmath>\begin{align*}
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\prod_{i=1}^{7}(a_i^7+1)&=\prod_{i=1}^{7}(a_i^3-1)\\
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&=\prod_{i=1}^{7}(a_i-1)\prod_{i=1}^{7}(a_i-\omega)\prod_{i=1}^{7}\left(a_i-\frac{1}{\omega}\right)\\
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&=-\prod_{i=1}^{7}(1-a_i)\prod_{i=1}^{7}(\omega-a_i)\prod_{i=1}^{7}\left(\frac{1}{\omega}-a_i\right)\\
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&=-f(1)f(\omega)f\left(\frac{1}{\omega}\right)\\
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&=-2(\omega^7-\omega^3+2)\left(\left(\frac{1}{\omega}\right)^7-\left(\frac{1}{\omega}\right)^3+2\right)\\
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&=-2(\omega-1+2)\left(\frac{1}{\omega}-1+2\right)\\
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&=-2(\omega+1)\left(\frac{1}{\omega}+1\right)\\
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&=-2\left(1+\left(\omega+\frac{1}{\omega}\right)+1\right)\\
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&=-2\left(1-1+1\right)=-2.
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\end{align*}</cmath>
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Thus, our answer is <math>(-2)^2 = \boxed{4}.</math>
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~SMO_Team
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==Solution 2==
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Let <math>a_1, a_2, \dots, a_7</math> be the solutions to <math>x^7 - x^3 + 2</math>. The set <math>a_1, a_2, \dots a_{14}</math> is just <math>2</math> duplicates of this set (all roots are double roots due to square = 0 condition), so we can just calculate the given product over these roots and then square the final result.
 +
 
 +
Since <math>(a_i)^7 = (a_i)^3 - 2</math> for all <math>1 \leq i \leq 7</math>, since these are roots to the equation, we have that <cmath>\prod_{i=1}^{7} (a_i)^7 + 1 = \prod_{i=1}^{7} (a_i)^3 - 1 = \prod_{i=1}^{7} (a_i-1)(a_i-w)(a_i-w^2)</cmath> where <math>w</math> is a third root of unity. Let <math>f(x) = x^7 - x^3 + 2 = \prod_{i=1}^{7} (x-a_i)</math> and break the above products into <math>3</math> separate products, we have that <cmath>\prod_{i=1}^{7} (a_i-1)(a_i-w)(a_i-w^2) = -\prod_{i=1}^{7} (1-a_i)(w-a_i)(w^2-a_1) = -f(1)\cdot f(w)\cdot f(w^2) = -2</cmath> Therefore, the answer is equal to the square of this product: <math>(-2)^2 = \boxed{4}</math>.
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 +
-Vivdax

Latest revision as of 14:29, 10 September 2025

Problem

Let $a_1, a_2, \dots a_{14}$ be the roots of $(x^7-x^3+2)^2=0$. Find the value of $\prod_{i=1}^{14} (a_i^7+1)$.

Solution 1

Let $a_1,a_2,\dots,a_7$ be the roots of \[f(x) = x^7-x^3+2 = \prod_{i=1}^7(x-a_i).\] It is easy to see that \[\prod_{i=1}^{14}(a_i^7+1) = \left(\prod_{i=1}^{7}(a_i^7+1)\right)^2.\] Now, we have $a_i^7-a_i^3+2 = 0\implies a_i^7+1 = a_i^3-1.$ Denote $1,\omega,\frac{1}{\omega}$ as the roots of $x^3-1,$ with $\omega^3 = 1.$ So, \begin{align*} \prod_{i=1}^{7}(a_i^7+1)&=\prod_{i=1}^{7}(a_i^3-1)\\ &=\prod_{i=1}^{7}(a_i-1)\prod_{i=1}^{7}(a_i-\omega)\prod_{i=1}^{7}\left(a_i-\frac{1}{\omega}\right)\\ &=-\prod_{i=1}^{7}(1-a_i)\prod_{i=1}^{7}(\omega-a_i)\prod_{i=1}^{7}\left(\frac{1}{\omega}-a_i\right)\\ &=-f(1)f(\omega)f\left(\frac{1}{\omega}\right)\\ &=-2(\omega^7-\omega^3+2)\left(\left(\frac{1}{\omega}\right)^7-\left(\frac{1}{\omega}\right)^3+2\right)\\ &=-2(\omega-1+2)\left(\frac{1}{\omega}-1+2\right)\\ &=-2(\omega+1)\left(\frac{1}{\omega}+1\right)\\ &=-2\left(1+\left(\omega+\frac{1}{\omega}\right)+1\right)\\ &=-2\left(1-1+1\right)=-2. \end{align*} Thus, our answer is $(-2)^2 = \boxed{4}.$

~SMO_Team

Solution 2

Let $a_1, a_2, \dots, a_7$ be the solutions to $x^7 - x^3 + 2$. The set $a_1, a_2, \dots a_{14}$ is just $2$ duplicates of this set (all roots are double roots due to square = 0 condition), so we can just calculate the given product over these roots and then square the final result.

Since $(a_i)^7 = (a_i)^3 - 2$ for all $1 \leq i \leq 7$, since these are roots to the equation, we have that \[\prod_{i=1}^{7} (a_i)^7 + 1 = \prod_{i=1}^{7} (a_i)^3 - 1 = \prod_{i=1}^{7} (a_i-1)(a_i-w)(a_i-w^2)\] where $w$ is a third root of unity. Let $f(x) = x^7 - x^3 + 2 = \prod_{i=1}^{7} (x-a_i)$ and break the above products into $3$ separate products, we have that \[\prod_{i=1}^{7} (a_i-1)(a_i-w)(a_i-w^2) = -\prod_{i=1}^{7} (1-a_i)(w-a_i)(w^2-a_1) = -f(1)\cdot f(w)\cdot f(w^2) = -2\] Therefore, the answer is equal to the square of this product: $(-2)^2 = \boxed{4}$.

-Vivdax

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