7. Let
$$
f(x)=x^{4}-6 x^{3}+26 x^{2}-46 x+65 \text {. }
$$

Let the roots of $f(x)$ be $a_{k}+i b_{k}$ for $k=1,2,3,4$. Given that the $a_{k}, b_{k}$ are all integers, find $\left|b_{1}\right|+\left|b_{2}\right|+\left|b_{3}\right|+\left|b_{4}\right|$.
Answer: 10
Solution: The roots of $f(x)$ must come in complex-conjugate pairs. We can then say that $a_{1}=a_{2}$ and $b_{1}=-b_{2} ; a_{3}=a_{4}$ and $b_{3}=-b_{4}$. The constant term of $f(x)$ is the product of these, so $5 \cdot 13=\left(a_{1}{ }^{2}+b_{1}{ }^{2}\right)\left(a_{3}{ }^{2}+b_{3}{ }^{2}\right)$. Since $a_{k}$ and $b_{k}$ are integers for all $k$, and it is simple to check that 1 and $i$ are not roots of $f(x)$, we must have $a_{1}{ }^{2}+b_{1}{ }^{2}=5$ and $a_{3}{ }^{2}+b_{3}{ }^{2}=13$. The only possible ways to write these sums with positive integers is $1^{2}+2^{2}=5$ and $2^{2}+3^{2}=13$, so the values of $a_{1}$ and $b_{1}$ up to sign are 1 and 2 ; and $a_{3}$ and $b_{3}$ up to sign are 2 and 3 . From the $x^{3}$ coefficient of $f(x)$, we get that $a_{1}+a_{2}+a_{3}+a_{4}=6$, so $a_{1}+a_{3}=3$. From the limits we already have, this tells us that $a_{1}=1$ and $a_{3}=2$. Therefore $b_{1}, b_{2}= \pm 2$ and $b_{3}, b_{4}= \pm 3$, so the required sum is $2+2+3+3=10$.