## Problem B1

Let $a_{n}, b_{n}$ be two sequences of integers such that: (1) $a_{0}=0, b_{0}=8$; (2) $a_{n+2}=2 a_{n+1}-a_{n}+2$, $b_{n+2}=2 b_{n+1}-b_{n}$, (3) $a_{n}{ }^{2}+b_{n}{ }^{2}$ is a square for $n>0$. Find at least two possible values for $\left(a_{1992}\right.$, $\mathrm{b}_{1992}$ ).

## Answer

(1992. 1996, 4. 1992+8), (1992. 1988, -4. 1992+8)


## Solution

$a_{n}$ satisfies a standard linear recurrence relation with general solution $a_{n}=n^{2}+A n+k$. But $a_{n}$ $=0$, so $k=0$. Hence $a_{n}=n^{2}+$ An. If you are not familiar with the general solution, then you can guess this solution and prove it by induction.

Similarly, $b_{n}=B n+8$. Hence $a_{n}{ }^{2}+b_{n}{ }^{2}=n^{4}+2 A n^{3}+\left(A^{2}+B^{2}\right) n^{2}+16 B n+64$. If this is a square, then looking at the constant and $n^{3}$ terms, it must be ( $\left.n^{2}+A n+8\right)$. Comparing the other terms, $\mathrm{A}=\mathrm{B}= \pm 4$.

