AIME 2026 II · 第 11 题

Solution 1

Notice that positive/negative values for $\alpha$, $\beta$, $\gamma$ give eight distinct values for $\alpha+\beta+\gamma$, so we conclude that for a choice of the three variables, $\alpha+\beta+\gamma=0$. (Testing other cases yields too few possible values.)

This is due to the fact that if we take either $\alpha,\beta,\gamma$ or $-\alpha,-\beta,-\gamma$, both sums are equal to $0$.

By Vieta, we get that

$$n-11=\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=(\alpha\beta+\beta\gamma+\gamma\alpha)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma)=(\alpha\beta+\beta\gamma+\gamma\alpha)^2$$ $$\frac{n}{6}=\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=-2(\alpha\beta+\beta\gamma+\gamma\alpha)=\implies \alpha\beta+\beta\gamma+\gamma\alpha=-\frac{n}{12}$$ Substituting, we get

$$n-11=\left(-\frac{n}{12}\right)^2=\frac{1}{144}n^2$$ Solving the quadratic gives positive integer solutions $n=12,\boxed{132}$.

~ eevee9406

Solution 2

We continue with the first solution with $\alpha + \beta + \gamma = 0$. Then, let $P(x)$ be the polynomial with roots $\alpha$, $\beta$, and $\gamma$ and let $Q(x)$ be the polynomial in the problem. Then,

$$P(x) = (x-\alpha) (x-\beta) (x-\gamma)$$ $$P(-x) = -(x + \alpha)(x + \beta)(x + \gamma).$$ From Vieta's, there exists complex $p$ and $q$ where $P(x) = x^3 + px + q$.

$$-P(x)P(-x) = Q(x^2)$$ $$-(x^3 + px + q)(-x^3 - px + q) = x^6 - \frac{n}{6} x^4 + (n-11)x^2 - 400,$$ $$x^6 + 2px^4 + p^2x^2 - q^2 = x^6 - \frac{n}{6} x^4 + (n-11)x^2 - 400.$$ Comparing coefficients, we need $2p = \frac{n}{6}$ and $p^2 = n-11$. Solving for $n$, we either $n = 12$ or $132$, but we need the bigger one, so the answer is $\boxed{132}$.

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