8. The number $a$ is a root of the equation $x^{11}+x^{7}+x^{3}=1$. Indicate all natural values of $n$ for which the equality $a^{4}+a^{3}=a^{n}+1$ holds.
Answer: Only 15.

Solution: Since $a$ is a root of the equation $x^{11} + x^{7} + x^{3} = 1$, then $a^{11} + a^{7} + a^{3} = 1$.

If we factor out $a^{3}$ from the left side, we get $a^{8} + a^{4} + 1$ - this is the expression $s^{2} + s + 1$ for $s = a^{4}$, which we have seen in the formula for the difference of cubes $s^{3} - 1 = (s - 1)(s^{2} + s + 1)$. Multiplying the equation by $a^{4} - 1$, we get $a^{3}(a^{12} - 1) = a^{4} - 1$, from which, after transformations, we obtain $a^{15} + 1 = a^{3} + a^{4}$.

It remains to note that $a \neq \pm 1$ and $a \neq 0$, since these numbers are not roots of the original equation. Therefore, all powers of $a$ are distinct, so nothing other than 15 is the answer.

Remark: The number $a$ exists: the value of $x^{11} + x^{7} + x^{3}$ at $x = 0$ is 0, i.e., less than 1, and at $x = 1$ it is 3, i.e., greater than 1. By the Intermediate Value Theorem, somewhere on the interval $[0,1]$ it must equal 1.

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