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Here's Le Monde puzzle #824:
Show that, for any integer \(y\), \[(\sqrt{3}-1)^{2y}+(\sqrt{3}+1)^{2y}\] is an integer multiple of a power of two.
Solution:
Consider \[f(n) = (-1+\sqrt{3})^{n}+(-1-\sqrt{3})^{n}\] and observe that the two bases are the roots of the quadratic \(x^2 + 2x - 2\), hence \( f(n) \) obeys the recursion \( x_{n+2} = -2 x_{n+1} + 2 x_n \) with \( x_0=2\) and \(x_1=-2 \). It follows that \( f(n) \) is always an even integer.
Show that, for any integer \(y\), \[(\sqrt{3}-1)^{2y}+(\sqrt{3}+1)^{2y}\] is an integer multiple of a power of two.
Solution:
Consider \[f(n) = (-1+\sqrt{3})^{n}+(-1-\sqrt{3})^{n}\] and observe that the two bases are the roots of the quadratic \(x^2 + 2x - 2\), hence \( f(n) \) obeys the recursion \( x_{n+2} = -2 x_{n+1} + 2 x_n \) with \( x_0=2\) and \(x_1=-2 \). It follows that \( f(n) \) is always an even integer.
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