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ysharifi的博客)
Problem. Given integers $a,b,c,d$, consider the sequence\[x_n:=\left(\frac{a+d+\sqrt{(a-d)^2+4bc}}{2}\right)^n + \left(\frac{a+d-\sqrt{(a-d)^2+4bc}}{2}\right)^n\]Show that $x_n$ is an integer for all integers $n\ge1$.
Solution. Consider the matrix $A:=\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in M_2(\mathbb{Z})$.
Let, as always, $I \in M_2(\mathbb{Z})$ be the identity matrix. We now find the eigenvalues of $A$. We have\[\det(xI-A)=\det \begin{pmatrix}x-a & -b \\ -c & x-d\end{pmatrix}=x^2-(a+d)x+ad-bc,\]and so the eigenvalues of $A$, which are the roots of the above quadratic polynomial, are\[\lambda_1=\frac{a+d+\sqrt{(a-d)^2+4bc}}{2}, \ \ \ \ \ \ \lambda_2= \frac{a+d-\sqrt{(a-d)^2+4bc}}{2}.\]Now, since all the entries of $A$ are integers, all the entries of $A^n$ are also integers for all integers $n \ge 1$. Thus the trace of $A^n$ is an integer for all $n \ge 1$. But the trace of $A^n$ is the sum of the eigenvalues of $A^n$ and the eigenvalues of $A^n$ are $\lambda_1^n, \lambda_2^n$. Thus $x_n=\lambda_1^n+\lambda_2^n$ is an integer for all $n \ge 1$.$ \ \Box$
Remark. Given integers $a,b$, it is clear that\[c_n:=\left(\frac{a+\sqrt{b}}{2}\right)^n+\left(\frac{a-\sqrt{b}}{2}\right)^n\]is a rational number for all integers $n \ge 1$. But $c_n$ is not always an integer; e.g., choose $a=1, b=0, \ n=2$. So, considering the above problem, we got a good question to think about: can we find all the integers $a,b$ for which $c_n$ is an integer for all integers $n \ge 1$?
Czhang271828
Posted 2022-12-23 19:58
