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If $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \in \mathbb{Z}$$ and $$\frac{b}{a} + \frac{c}{b} + \frac{a}{c} \in \mathbb{Z}.$$

For any natural $n$, is the following true? $$\frac{a^n}{b^n}+\frac{b^n}{c^n}+\frac{c^n}{a^n} \in \mathbb{Z}$$

P.S: $a,b,c$ can be complex.

Full disclosure: I have a trick in my mind for generalizing problems of this type and this problem is motivated from the two term problem (shared in the link). I want to see a proof similar to the one I shared in the link.

Thanks.

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Isomorphism
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1 Answers1

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Let $p,q,r$ be equal to $\frac ab, \frac bc, \frac ca$, then $pqr=1$ and $p,q,r$ are the roots of the cubic $$f(x)=(x-p)(x-q)(x-r)=0=x^3-(p+q+r)x^2+(pq+qr+rp)x-1$$

Now $pq+qr+rp=\frac ac+\frac ba+\frac cb$ and $p+q+r$ are given to be integers, so $f(x)$ has integer coefficients, say $f(x)=x^3-sx^2+tx-1$.

Let $A_n=p^n+q^n+r^n$

Note that $f(p)=0$ so that $p^nf(p)=0$ so that $$0=p^nf(p)+q^nf(q)+r^nf(r)=A_{n+3}-sA_{n+2}+tA_{n+1}-A_n$$ and we have the linear recurrence with integer coefficients $$A_{n+3}=sA_{n+2}-tA_{n+1}+A_n$$

Since $A_{-1}$ and $A_1$ are given to be integers, and $A_0=3$, we have the basis for an induction, starting with $n=-1$ and the statement is true.

Mark Bennet
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  • :) This was my generalization. In fact, any symmetric polynomial is a polynomial in elementary symmetric polynomials and here the elementary symmetric polynomials are integers, so every symmetric polynomials in those fractions are integers.

    I want to obtain a problem where proving an identity is an integer is hard when we use these tricks. In this way I can motivate Gauss symmetric polynomial theorem.

    Thanks for the answer :)

     – Isomorphism Sep 18 '14 at 15:29