1998 IMO Shortlist Problems/N1: Difference between revisions
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>" |
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== 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, ![\[ab^2+b+7|a^2b+a+b\]](http://latex.artofproblemsolving.com/5/1/4/514740d18c9510d3ae694d84febb9406efd6b3ce.png)
Solution
We have the following divisibility relations, ![\[ab^2+b+7|a^2b+a+b|b(a^2b+a+b)=a^2b^2+ab+b^2\]](http://latex.artofproblemsolving.com/e/b/2/eb27791c9e93fd6fe722ce6b455e6b8602f43c61.png)
![\[ab^2+b+7|a(ab^2+b+7)=a^2b^2+ab+7a\]](http://latex.artofproblemsolving.com/8/4/5/845cb91b587505a7773a26a9aad6b80d873aa9e1.png)
Subtracting, ![\[ab^2+b+7| \;|b^2-7a|\]](http://latex.artofproblemsolving.com/8/d/d/8ddbe4554847a8fceadd43c9e80e4ecf8d150703.png)
If 
, we may have ![\[a+8|7a-1\]](http://latex.artofproblemsolving.com/e/d/9/ed95e155f9d5b641883333ba130c37851e218ebf.png)
Otherwise, we would have ![\[|b^2-7a|<ab^2+b+7\]](http://latex.artofproblemsolving.com/8/a/c/8ac8947648e59ced8697b2673de55062e1889f1c.png)
In the first case, ![\[a+8|7a+56\]](http://latex.artofproblemsolving.com/6/0/d/60d007f9edcbb58083cbad9570b5272126412a92.png)
This gives, ![\[a+8|57\]](http://latex.artofproblemsolving.com/7/c/8/7c8daf385fdfad078bcf8f56f00339a4daf0ad4e.png)
Since, 
and 
, we must have 
or 
which yields the solutions 
.
In the second case, 
or, ![\[b^2=7a\implies 7|b\]](http://latex.artofproblemsolving.com/d/0/8/d087b6111c8d887d62399693554ae60aae66d791.png)
Say, 
. Then 
. Checking we find that it is indeed a solution.
Thus, all solutions are ![\[(a,b)=(11,1),(49,1),(7k^2,7k).\]](http://latex.artofproblemsolving.com/b/d/1/bd11f0b5c3022f0b9661c258e04c7b6f3329307d.png)
