Difference between revisions of "1992 OIM Problems/Problem 4"
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== Solution == | == Solution == | ||
− | + | First we find the non-recursive form of this with unknown <math>a_1</math> and <math>b_1</math>: | |
− | {{ | + | <math>a_n=na_1+n(n+1))</math>, and <math>b_n=nb_1-8(n-1)</math> |
+ | |||
+ | Let <math>A=a_1-1</math>, and <math>B=b_1-8</math> | ||
+ | |||
+ | <math>a_n=n^2+An</math>, and <math>b_n=Bn+8</math> | ||
+ | |||
+ | <math>a_n^2+b_n^2=(n^2+An)^2+(Bn+8)^2</math> | ||
+ | |||
+ | <math>a_n^2+b_n^2=n^4+2An^3+(A^2+B^2)n^2+16Bn+8^2=S^2</math> | ||
+ | |||
+ | Let <math>S^2=(n^2+Kn+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>, and <math>A=B=\pm 4</math> | ||
+ | |||
+ | Substituting we have: | ||
+ | |||
+ | <math>a_n=n^2\pm 4n</math>, and <math>b_n=\pm 4n+8</math> | ||
+ | |||
+ | Thus | ||
+ | |||
+ | <math>(a_{1992},b_{1992})=(1992^2 + 4\cdot 1992,4\cdot 1992+8)</math>, or <math>(1992^2 - 4\cdot 1992,-4\cdot 1992+8)</math> | ||
+ | |||
+ | <math>(a_{1992},b_{1992})=(3976032,7976)</math>, or <math>(3960096,-7960)</math> | ||
+ | |||
+ | * 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 10 on this one. I don't remember what I did. | ||
+ | |||
+ | ~Tomas Diaz. ~orders@tomasdiaz.com | ||
+ | |||
+ | |||
+ | {{alternate solutions}} | ||
== See also == | == See also == | ||
+ | [[OIM Problems and Solutions]] | ||
+ | |||
https://www.oma.org.ar/enunciados/ibe7.htm | https://www.oma.org.ar/enunciados/ibe7.htm |
Latest revision as of 08:42, 23 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 , and
Substituting we have:
, and
Thus
, or
, or
- 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 10 on this one. I don't remember what I did.
~Tomas Diaz. ~orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.