Suppose that $x$ and $y$ are complex numbers such that \[\frac{x^{n}-y^{n}}{x-y}\] are integers for some four consecutive positive integers $n$. Prove that it is an integer for all positive integers $n$.
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Tags: algebra, polynomial, induction, complex numbers, Miscellaneous Problems, pen
TTsphn
22.10.2007 20:55
$ c = xy,b = x + y$ We has $ a_{n + 2} = ba_{n + 1} - ca_n$ We have : $ a_{n + 1}^2 - a_na_{n + 2} = c^n$ We prove by induction. Call $ a_m,a_{m + 1},a_{m + 2},a_{m + 3}$ is four consective integer in this sequence. $ c^m = a_{m + 1}^2 - a_ma_{m + 2}$ $ c^{m + 1} = a_{m + 2}^2 - a_{m + 1}a_{m + 3}$ Imply that $ c\in Q$ Because $ c^m\in Z$ so $ c\in Z$ Suppose $ c=\frac{p}{q}$ where $ \gcd(p,q)=1$ so $ q^m|p^m\Rightarrow q=1$ so $ c\in Z$ $ b = \frac {a_{m + 2} - ca_m}{a_{m + 1}}\Rightarrow b\in Q$ The general of sequence imply that b is a rational root of a monic Polynomial .(Because $ a_n = f_n(b)$) so $ b\in Z$ imply that $ b,c\in Z$ We has $ a_1 = 1,a_2 = b$ so $ a_n\in Z,\forall n\in N$
Peter
23.10.2007 01:00
TTsphn wrote: We has $ a_{n + 2} = ba_{n + 1} + ca_n$ Really? TTsphn wrote: $ a_{n + 1}^2 - a_na_{n + 2} = c^n$ How?
TTsphn
24.10.2007 15:46
Peter wrote: TTsphn wrote: We has $ a_{n + 2} = ba_{n + 1} + ca_n$ Really? TTsphn wrote: $ a_{n + 1}^2 - a_na_{n + 2} = c^n$ How? Sorry I have edit it.
Peter
24.10.2007 16:33
I figured it should be a minus, but I still don't see how that leads you to $ a_{n + 1}^2 - a_na_{n + 2} = c^n$?
TTsphn
24.10.2007 16:44
Proof We will induction on N $ a_1 = 1$ $ a_2 = x + y$ $ a_3 = x^2 + xy + y^2$ so $ a_{2}^2 - a_1a_3 = c$ Now we prove that $ a_{n + 2}^2 - a_{n + 1}a_{n + 3} = c(a_{n + 1}^2 - a_na_{n + 2}$ $ \Leftrightarrow a_{n + 2}(a_{n + 2} + ca_n) = a_{n + 1}(ca_{n + 1} + a_{n + 3})$ But $ a_{n + 2} + ca_n = ba_{n + 1}$ $ a_{n + 3} + ca_n = ba_{n + 2}$ So the equality is true, Imply that $ a_{n + 2}^2 - a_{n + 1}a_{n + 3} = c(a_{n + 1}^2 - a_na_{n + 2}) = .. = c^{n + 1}$ So it true.
Wolstenholme
14.06.2013 02:19
How did you get from b is rational to b is an integer? I understand that if it is the root of a monic polynomial then it must be an integer, but what monic polynomial?
tpdtthltvp
23.06.2018 19:12
TTsphn wrote: $ c = xy,b = x + y$ We has $ a_{n + 2} = ba_{n + 1} - ca_n$ We have : $ a_{n + 1}^2 - a_na_{n + 2} = c^n$ We prove by induction. Call $ a_m,a_{m + 1},a_{m + 2},a_{m + 3}$ is four consective integer in this sequence. $ c^m = a_{m + 1}^2 - a_ma_{m + 2}$ $ c^{m + 1} = a_{m + 2}^2 - a_{m + 1}a_{m + 3}$ Imply that $ c\in Q$ Because $ c^m\in Z$ so $ c\in Z$ Suppose $ c=\frac{p}{q}$ where $ \gcd(p,q)=1$ so $ q^m|p^m\Rightarrow q=1$ so $ c\in Z$ $ b = \frac {a_{m + 2} - ca_m}{a_{m + 1}}\Rightarrow b\in Q$ The general of sequence imply that b is a rational root of a monic Polynomial .(Because $ a_n = f_n(b)$) so $ b\in Z$ imply that $ b,c\in Z$ We has $ a_1 = 1,a_2 = b$ so $ a_n\in Z,\forall n\in N$ What is the monic polynomial?