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We have sequence

$T_1=1,T_2=3,T_n=2T_{n-1}+T_{n-2}$, where $n\ge3$

I need to prove that for every natural number $n\ge2$, GCD$(T_n,T_{n-1})=1$

Any ideas how to prove it? I noticed that first five member of this sequence is a prime number, except $T_6=99$, and then there goes primes again. How to prove it in general?

Spameris
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  • See https://math.stackexchange.com/a/214891/42969 for a more general result. – Martin R Oct 13 '19 at 18:40
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    If $p$ divides $T_n$ and $T_{n-1}$, then recurrence gives $p \mid T_{n-2}$, repeat the process to reach $p\mid T_1=1$. – Sil Oct 13 '19 at 18:58
  • You can also show that $T_n^2-T_{n-1}T_{n+1}=2,(-1)^n$, so $\gcd(T_n,T_{n-1})$ must divide $2$, and since each $T_n$ is odd, the claim follows. (However, this equality is much stronger than what is needed to solve the problem.) – Batominovski Oct 13 '19 at 19:51

2 Answers2

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Hint for induction: $\gcd(x,y)=\gcd(x+y,y)$.

(This is true because $x,y$ and $x+y,y$ have the same sets of common divisors!)

anon
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We have $$ \begin{pmatrix}T_{n+2} \\ T_{n+1} \end{pmatrix} =\begin{pmatrix}2&1\\1&0\end{pmatrix} \begin{pmatrix}T_{n+1} \\ T_{n} \end{pmatrix} $$ The matrix has determinant $-1$ and so is invertible over the integers.

Therefore, the common divisors of $T_{n+2}, T_{n+1}$ are exactly the same as the common divisors of $T_{n+1}, T_{n}$. The result follows by induction since $\gcd(T_1,T_0)=1$.

lhf
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