Difference between revisions of "1992 OIM Problems/Problem 4"
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<math>S^2=n^4+2Kn^3+(16+K^2)n^2+16Kn+8^2</math> | <math>S^2=n^4+2Kn^3+(16+K^2)n^2+16Kn+8^2</math> | ||
+ | From the coefficient in front of <math>n^3</math> we find <math>2K=2A</math> thus <math>A=K</math> | ||
+ | |||
+ | From the coefficient in front of <math>n</math> we find <math>16K=16B</math> thus <math>B=K</math>, and <math>A=B=K</math> | ||
+ | |||
+ | 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> | ||
Revision as of 20:23, 19 December 2023
Problem
Let and be two sequences of integers that verify the following conditions:
i. ,
ii. For all , ,
iii. is a perfect square for all
Find at least two values of pair .
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
First we find the non-recursive form of this with unknown and :
, and
Let , and
, and
Let
From the coefficient in front of we find thus
From the coefficient in front of we find thus , and
From the coefficient in front of we have:
therefore , thus
- 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.