Solutions to $a,\ b,\ c,\ \frac{a}{b}+\frac{b}{c}+\frac{c}{a},\ \frac{b}{a} + \frac{c}{b} + \frac{a}{c} \in \mathbb{Z}$

Suppose that $\displaystyle a,b,c,\frac{a}{b}+\frac{b}{c}+\frac{c}{a},\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \in \mathbb Z$. Consider polynomial $$P(x)=\left(x-\frac{a}{b}\right)\left(x-\frac{b}{c}\right)\left(x-\frac{c}{a}\right) = x^3-\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)x^2+\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)x-1.$$ Its coefficients are integers. Since the leading coefficient is $1$, all rational roots of $P$ are integers. Since the constant term is $-1$, it follows that all integer roots of $P$ are $1$ or $-1$ (they must divide the constant term). Since $\dfrac ab, \dfrac bc, \dfrac ca$ are rational roots of $P$, it follows that $\dfrac ab, \dfrac bc, \dfrac ca \in \{-1,1\}$.

Let $(a,b,c)$ satisfy the requirements. Let $(a,b,c)$ are coprime. Then $abc$ divides both $a^2c + b^2a + c^2b$ and $a^2b+b^2c+c^2a$.

Let $p$ be a prime factor of $a$. Let $d$ be the largest number such that $p^d$ divides $a$. Then $p$ divides $b^2c$ and $c^2b$. Assume $p$ divides $b$ (and does not divide $c$).

Since $p^{d+1}$ divides $a^2$, $ab$, and $abc$, where the latter divides $a^2c + b^2a + c^2b$, it follows $p^{d+1}$ divides $b$. This in turn implies that $p^{d+1}$ divides $a^2b+b^2c+c^2a$. Now, $p$ does not divide $c$ by assumption of coprimality, hence $p^{d+1}$ divides $a$, a contradiction to the maximality of $d$.

Hence, none of $a,b,c$ has a prime factor. So all these numbers are equal $\pm1$.