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We define $T_k = x^k + \frac{1}{x^k}$,for some non zero real number $x$. Then one can look at the algebra generated by $T_k$. Its simple to see that they satisfy

$$T_k T_l = T_{k+l} + T_{|k-l|} $$ Also we have that $T_{-k} = T_{k}, T_0 =2$.

It is simple to see via this relation that $T_1$ generates all $T_k$.

We are given that there is a choice of $x \in \mathbb{R}$ such that $ \exists m,n \in \mathbb{N}$ and $T_m,T_n \in \mathbb{Q}$. The question is to find the sufficient conditions that $m,n$ satisfy to imply that all $T_k$ are forced to be rational. ( this would translate to producing $T_1$ as a linear combination in the set generated by $T_n,T_m$)

As an example consider that $T_3,T_5 \in \mathbb{Q}$, then $T_3 T_3 = T_6 +2, T_5 T_5 = T_{10} +2, T_6 T_6 = T_{12} +2$, which tells us that $T_6,T_{10},T_{12} \in \mathbb{Q}$. Subtracting $T_{10} T_2 = T_{12} +T_8$ from $T_3 T_5 = T_2+T_8$ gives us $T_3 T_5 +T_{12} = T_2(1+T_{10})$, which gives us that $T_2$ is rational, and $T_1.T_2 = T_3 + T_1$, which tells that $T_1$ is rational, hence all $T_k$ are rational.

(p.s.: I feel that a necessary and sufficient condition is to have that $gcd(m,n)=1$.)

Sagars
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1 Answers1

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Further to Martin R's comment.

In the case that $m,n$ are co-prime one can reduce to the case where they differ by $1$ by taking multiples which differ by 1 (these multiples still give rational T) and using the answers to this question

If $m,n$ have a common multiple $d\gt 1$ then set $x=\sqrt[d] 2$ to give an example where $T$ is rational only at multiples of $d$.

Of course in the case where $m,n$ have a common multiple $d\gt 1$ it is still possible that $T_1$ can be rational. Just choose the $T_m, T_n$ which are generated by your chosen value for $T_1$

Mark Bennet
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