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

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TopNewestWe 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. – Bogdan Simeonov · 3 years ago

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– Eddie The Head · 3 years ago

Nice ..I also solved in the same way....Log in to reply