Find all rational terms of sequence defined by formula $ a_n=\sqrt{\frac{9n-2}{n+1}}, n \in N $
Problem
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Tags: Diophantine equation
Thjch Ph4 Trjnh
20.05.2010 17:52
Syler wrote: Find all rational terms of sequence defined by formula $ a_n=\sqrt{\frac{9n-2}{n+1}}, n \in N $ We have: $(9n-2,n+1)=(n+1,11)$. *) If $n+1=11k$, then $9k-1,k$ are perfect squares. $9k-1=a^2,k=b^2$. Then $a^2-9b^2=-1$, Pell's equation. *) If $(n+1,11)=1$ then $9n-2=a^2,n+1=b^2$. $\Rightarrow a^2-9b^2=-11$.
Jason-Kountou
21.05.2010 17:53
It suffices to find all k-th terms of the sequence that satisfy $ \frac{9k-2}{k+1} = \frac{p^2}{q^2}$ where $ p,q \in \mathbb{Z}, (p,q) = 1$.
This is written as $ 9q^2 - \frac{11q^2}{k+1} = p^2 \Leftrightarrow 11q^2 = (3q-p)(3q+p)(k+1)$
From there it is obvious that either $ q^2 \mid 1$ and so the problem becomes "find all integer terms of the sequence" or $ q^2 \mid k +1 $.
In the first case this becomes $9 - \frac{11}{k+1} = p^2$ so $ k = 0, 10$, none of which satisfy the conditions.
In the second case we write $ k = mq^2 -1$ and continue to find the pairs of p,q such that $ 11 = (3q-p)(3q+p)m$
Since m > 0 we split into cases:
(i) $ m =11, 3q-p = 3q+p = 1 $ from which we get $ p = 0, q = \frac{1}{3} $ (contradiction) (same if we have 3q - p = 3q + p = -1)
(ii) $ m = 1, 3q - p = 1, 3q + p = 11$ from which we get $ q = 2, p = 5$ and $ k = 3$ (pairs (p,q)=(-5,-2)(5,-2)(3,-2) are also acceptable)
So the third term is the only rational term in the sequence.
The above solution kinda comes in contradiction to mine, which is why I'm confused. Perhaps I've made a mistake somewhere.
mavropnevma
21.05.2010 19:14
No mistake, Jason! The "solution" previous to yours calls $a^2 - 9b^2 = -1$ a Pell equation, which it isn't, since $9$ is a perfect square. In fact that has no solutions (check it modulo $3$). The second relation $a^2 - 9b^2 = -11$ is neither a Pell equation; since it writes $(a-3b)(a+3b) = -11$, it only has solution $a=5$, $b=2$, which means, as you said, only the third term is integer.