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Taken from MIT's OpenCourseWare site for Discrete Math:

We define the following recurrence for $n ≥ 0$: $$T_{n+2} = T_{n+1} + 2T_{n}$$ where, $T_{0} = T_{1} = 1$

(a) Prove by induction that: $T_{n}$ is odd for $n ≥ 0.$

You do not need to solve the recurrence for this.

(b) Prove by induction that: $$\gcd(T_{n+1}, T_{n}) = 1 \qquad\forall n ≥ 0.$$ You may assume that $T_{n}$ is odd for all $n$.

You do not need to solve the recurrence for this.

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chzos
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1 Answers1

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$(a)$ Let $T_n$ is odd for $n=m+1$

Then $\displaystyle T_{m+2}=T_{m+1}+2T_m$ which is odd as for integer $\displaystyle T_m, T_{m+2}-T_{m+1}$ will be even i.e., $\displaystyle T_{m+2},T_{m+1}$ will have same parity

$(b)$ $$(T_{n+2},T_{n+1})=(T_{n+1}+2T_n,T_{n+1})=(2T_n,T_{n+1})=(T_n,T_{n+1})$$ as $T_{n+1}$ is odd

$$\implies(T_{n+2},T_{n+1})=(T_{n+1},T_n)$$ which reminds me of Prove that any two consecutive terms of the Fibonacci sequence are relatively prime

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lab bhattacharjee
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