Difference between revisions of "1998 IMO Shortlist Problems/N1"

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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>
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== Solution ==
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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>
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<cmath>ab^2+b+7|a(ab^2+b+7)=a^2b^2+ab+7a</cmath>
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Subtracting, <cmath>ab^2+b+7|  \;|b^2-7a|</cmath>
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If <math>b=1</math>, we may have <cmath>a+8|7a-1</cmath>
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Otherwise, we would have <cmath>|b^2-7a|<ab^2+b+7</cmath>
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In the first case, <cmath>a+8|7a+56</cmath>
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This gives, <cmath>a+8|57</cmath>
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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>.
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In the second case, <math>b^2-7a=0</math> or, <cmath>b^2=7a\implies 7|b</cmath>
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Say, <math>b=7k</math>. Then <math>a=7k^2</math>. Checking we find that it is indeed a solution.
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Thus, all solutions are <cmath>(a,b)=(11,1),(49,1),(7k^2,7k).</cmath>

Latest revision as of 04:20, 16 August 2011

Problem

Find all pairs of positive integers $(a,b)$ such that, \[ab^2+b+7|a^2b+a+b\]


Solution

We have the following divisibility relations, \[ab^2+b+7|a^2b+a+b|b(a^2b+a+b)=a^2b^2+ab+b^2\] \[ab^2+b+7|a(ab^2+b+7)=a^2b^2+ab+7a\] Subtracting, \[ab^2+b+7|   \;|b^2-7a|\]

If $b=1$, we may have \[a+8|7a-1\] Otherwise, we would have \[|b^2-7a|<ab^2+b+7\] In the first case, \[a+8|7a+56\] This gives, \[a+8|57\] Since, $a+8>3$ and $57=3\cdot19$, we must have $a+8=19$ or $57$ which yields the solutions $(a,b)=(11,1),(49,1)$.

In the second case, $b^2-7a=0$ or, \[b^2=7a\implies 7|b\] Say, $b=7k$. Then $a=7k^2$. Checking we find that it is indeed a solution. Thus, all solutions are \[(a,b)=(11,1),(49,1),(7k^2,7k).\]

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