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# Indian Regional Mathematical olympiad problem

Let $$\alpha$$ and $$\beta$$ be two roots of the quadratic equation $$x^{2}+mx-1 = 0$$,where $$m$$ is an odd integer.Let $$\lambda_n = \alpha^{n}+\beta^{n}$$ for $$n\ge 0$$.Prove that for $$n \ge 0$$,

(a) $$\lambda_n$$ is an integer.

(b)$$gcd( \lambda_n ,\lambda_{n+1}) = 1$$

Note by Eddie The Head
3 years, 2 months ago

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We get the relation $$\lambda_{n+1}=-m.\lambda_n+\lambda_{n-1}$$.

Since $$\lambda_0=2,\lambda_1=-m$$, it follows that $$(\lambda_0,\lambda_1)=1$$, because m is an odd integer.Also, since the first two terms are integers and we got a recurrence relation with integer coefficients, it follows that $$\lambda_n$$ must be an integer.

Firstly,$$(\lambda_0,\lambda_1)=1$$.Let $$(\lambda_n,\lambda_{n-1})=1$$.Then let's assume $$(\lambda_{n+1},\lambda_n)=d,d>1$$.Then, using the relation we would get that $$d|\lambda_{n-1}$$, which is clearly a contradiction, so we have proved it by induction. · 3 years, 2 months ago

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Nice ..I also solved in the same way.... · 3 years, 2 months ago

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