3. Find the number of triples of natural numbers $a, b$, and $c$ not exceeding 2017 such that the polynomial $x^{11} + a x^{7} + b x^{3} + c$ has a rational root.

Trunov K.V.
Answer: 2031120

Solution: Since all coefficients of the polynomial are natural and the leading coefficient is 1, any rational root must be an integer.

Notice that for $x \geq 0$, $x^{11} + a x^{7} + b x^{3} + c \geq c \geq 1$. Therefore, there are no integer roots for $x \geq 0$. If $x \leq -2$, then $x^{11} + a x^{7} + b x^{3} + c \leq -2^{11} - 2^{7} a - 2^{3} b + c \leq -2^{11} - 2^{7} - 2^{3} + 2017 < 0$, so there are no integer roots for $x \leq -2$. Therefore, only $x = -1$ can be an integer root of the polynomial $x^{11} + a x^{7} + b x^{3} + c$, and $x = -1$ is a root of the polynomial if the condition $-1 - a - b + c = 0 \Rightarrow c = 1 + a + b$ is satisfied. Since $a$, $b$, and $c$ are natural numbers and do not exceed 2017, $a$ can take values from 1 to 2015. For $a = 1$, $b$ can take values from 1 to 2015; for $a = 2$, $b$ can take values from 1 to 2014; ..., for $a = 2015$, $b$ can take only the value 1. Therefore, the total number of such triples is $1 + 2 + 3 + 4 + \ldots + 2015 = \frac{1 + 2015}{2} \cdot 2015 = 2031120$.

## Recommendations for checking.

Proved that all rational roots are integers - 1 point.

Proved that there are no integer roots for $x \geq 0$ - +1 point.

Proved that there are no integer roots for $x \leq -2$ - +3 points.