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

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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>.
 
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 19:59, 2 May 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

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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