How can I prove that $F_{2n}$ is divisible by $F_n$ in the Fibonacci sequence?
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Do you mean " is divisible by"? – Michael McGovern Feb 27 '18 at 22:23
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This question, or variants, seems to be coming up a lot over the past few days. Anyway, here is a reference. – lulu Feb 27 '18 at 22:23
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1In fact, $F_{2n} = F_n (F_{n+1} + F_{n-1})$. (My favorite proof of this: if $A = \begin{bmatrix} 1 & 1 \ 1 & 0 \end{bmatrix}$, then $A^n = \begin{bmatrix} F_{n+1} & F_n \ F_n & F_{n-1} \end{bmatrix}$. Now write $A^{2n} = (A^n)^2$ and compare the two sides.) – Daniel Schepler Feb 27 '18 at 22:29
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Since $(x^n-y^n)\mid (x^{2n}-y^{2n})$, it is a straightforward consequence of the closed formula for Fibonacci numbers. Actually $F_{2n}=L_n F_n$ where $L_n$ is the $n$-th Lucas number. – Jack D'Aurizio Feb 27 '18 at 22:30
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@JackD'Aurizio I think it's a bit more complex than that, since $x$ and $y$ are irrational. You would still be left with showing $x^n + y^n$ is an integer. – Daniel Schepler Feb 27 '18 at 22:32
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@DanielSchepler which is easy to show – rtybase Feb 27 '18 at 22:34
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@DanielSchepler: $x,y$ are conjugated roots for a monic quadratic polynomial with integer coefficients, so $x^n+y^n\in\mathbb{Z}$. Or, simply, that is true for $n=0,n=1$ and $H_{n}=x^n+y^n$ fulfills $H_{n+2}=H_{n+1}+H_n$. – Jack D'Aurizio Feb 27 '18 at 22:34
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Granted, it's not hard to show if you know about solutions to recurrences, or about symmetric polynomials; but otherwise, on its own as a problem out of the blue, it wouldn't be obvious. – Daniel Schepler Feb 27 '18 at 22:36