Difference between revisions of "1998 IMO Shortlist Problems/N1"
(Created page with "== Problem == Find all pairs of positive integers <math>(a,b)</math> such that, <cmath>ab^2+b+7|a^2b+a+b</cmath>") |
(→Problem) |
||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
Find all pairs of positive integers <math>(a,b)</math> such that, <cmath>ab^2+b+7|a^2b+a+b</cmath> | Find all pairs of positive integers <math>(a,b)</math> such that, <cmath>ab^2+b+7|a^2b+a+b</cmath> | ||
+ | |||
+ | |||
+ | |||
+ | == Solution == | ||
+ | We have the following divisibility relations, <cmath>ab^2+b+7|a^2b+a+b|b(a^2b+a+b)=a^2b^2+ab+b^2</cmath> | ||
+ | <cmath>ab^2+b+7|a(ab^2+b+7)=a^2b^2+ab+7a</cmath> | ||
+ | Subtracting, <cmath>ab^2+b+7| \;|b^2-7a|</cmath> | ||
+ | |||
+ | If <math>b=1</math>, we may have <cmath>a+8|7a-1</cmath> | ||
+ | Otherwise, we would have <cmath>|b^2-7a|<ab^2+b+7</cmath> | ||
+ | In the first case, <cmath>a+8|7a+56</cmath> | ||
+ | This gives, <cmath>a+8|57</cmath> | ||
+ | Since, <math>a+8>3</math> and <math>57=3\cdot19</math>, we must have <math>a+8=19</math> or <math>57</math> which yields the solutions <math>(a,b)=(11,1),(49,1)</math>. | ||
+ | |||
+ | In the second case, <math>b^2-7a=0</math> or, <cmath>b^2=7a\implies 7|b</cmath> | ||
+ | Say, <math>b=7k</math>. Then <math>a=7k^2</math>. Checking we find that it is indeed a solution. | ||
+ | Thus, all solutions are <cmath>(a,b)=(11,1),(49,1),(7k^2,7k).</cmath> |
Latest revision as of 03:20, 16 August 2011
Problem
Find all pairs of positive integers such that,
Solution
We have the following divisibility relations, Subtracting,
If , we may have Otherwise, we would have In the first case, This gives, Since, and , we must have or which yields the solutions .
In the second case, or, Say, . Then . Checking we find that it is indeed a solution. Thus, all solutions are