Difference between revisions of "1992 OIM Problems/Problem 4"

Revision as of 20:24, 19 December 2023

Let $(a_n)$ and $(b_n)$ be two sequences of integers that verify the following conditions:

i. $a_0 = 0$ , $b_0 = 8$

ii. For all $n \geq 0$ , $a_{n+2}=2a_{n+1}-a_{n}+2$ , $b_{n+2}=2b_{n+1}-b_{n}$

iii. $a_{n}^{2}+b_{n}^{2}$ is a perfect square for all $n\ge 0$

Find at least two values of pair $(a_{1992},b_{1992})$ .

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

First we find the non-recursive form of this with unknown $a_1$ and $b_1$ :

$a_n=na_1+n(n+1))$ , and $b_n=nb_1-8(n-1)$

Let $A=a_1-1$ , and $B=b_1-8$

$a_n=n^2+An$ , and $b_n=Bn+8$

$a_n^2+b_n^2=(n^2+An)^2+(Bn+8)^2$

$a_n^2+b_n^2=n^4+2An^3+(A^2+B^2)n^2+16Bn+8^2=S^2$

Let $S^2=(n^2+Kn+8)^2$

$S^2=n^4+2Kn^3+(16+K^2)n^2+16Kn+8^2$

From the coefficient in front of $n^3$ we find $2K=2A$ thus $A=K$

From the coefficient in front of $n$ we find $16K=16B$ thus $B=K$ , and $A=B=K$

From the coefficient in front of $n^2$ we have:

$(16+K^2)=(A^2+B^2)=(K^2+K^2)$ therefore $K^2=16$ , thus $K= \pm 4$ , and $A=B=\pm 4$

Note. I actually competed at this event in Venezuela when I was in High School representing Puerto Rico. I think I got like 2 or 3 points out of 1 on this one. I don't remember what I did.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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	From the coefficient in front of <math>n^2</math> we have:		From the coefficient in front of <math>n^2</math> we have:

−	<math>(16+K^2)=(A^2+B^2)=(K^2+K^2)</math> therefore <math>K^2=16</math>, thus <math>K= \pm 4</math>	+	<math>(16+K^2)=(A^2+B^2)=(K^2+K^2)</math> therefore <math>K^2=16</math>, thus <math>K= \pm 4</math>, and <math>A=B=\pm 4</math>