<s>[INST] To show that the variables in the system are same in magnitude I am stuck with this interesting problem,
If for non-negative integers $a, b, \text{and} c$, $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$ are both integers then show that $|a|=|b|=|c|$.
I could only make a few observations
1) Because the system is symmetric wrt. $a, b, c$ as the two expressions swap, when any two variables are swapped. So, we can assume that $a \geq b+1$ and $a \geq c+2$ as assuming equality of any two makes the problems trivial. 
2) Adding and subtracting the two equations give $\frac{(a+b)(b+c)(c+a)}{abc}$ and $\frac{(a-b)(b-c)(c-a)}{abc}$ are both integers.
Also, $\frac{(ab+bc+ca)(a+b+c)}{abc}$ is an integer.
Also,  seeing that the expressions involved in 2) are close to the elementary symmetry polynomials, I considered the polynomial $P(x)=(x-a)(x-b)(x-c)$
I think the above observations are useful for the problem but I have failed to build on them so someone please help me. Please!! By the way, I am in high school and don't know any college level stuff.
 [/INST] Let $m=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $n=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$. Then $\frac{a}{b},\frac{b}{c},\frac{c}{a}$ are the roots of $x^3-mx^2+nx-1=0$.
Hence we have $a^3-ma^2b+nab^2-b^3=0$, where $a,b,m,n$ are all integers, with $a,b$ positive. Let $p^r$ be the highest power of $p$ dividing $a$, and $p^s$ be the highest power of $p$ dividing $b$. Suppose $r>s$. Then the highest power of $p$ dividing $a^3-ma^2b+nab^2$ is at least $r+2s$, whereas the highest power of $p$ dividing $b^3$ is $3s<r+2s$. Contradiction. So we must have $r\le s$. Similarly $s\le r$, so $r=s$. This is true for any prime $p$, so we must have $a=b$. 
Similarly, $b=c$.
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