Difference between revisions of "1994 IMO Problems/Problem 4"
m (wikify) |
m (→Solution) |
||
(2 intermediate revisions by one other user not shown) | |||
Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
− | + | Suppose <math>\frac{n^3+1}{mn-1}=k</math> where <math>k</math> is a positive integer. Then <math>n^3+1=(mn-1)k</math> and so it is clear that <math>k\equiv -1\pmod{n}</math>. So, let <math>k=jn-1</math> where <math>j</math> is a positive integer. Then we have <math>n^3+1=(mn-1)(jn-1)=mjn^2-(m+j)n+1</math> which by cancelling out the <math>1</math>s and dividing by <math>n</math> yields <math>n^2=mjn-(m+j)\Rightarrow n^2-mjn+m+j=0</math>. The equation <math>x^2-mjx+m+j=0</math> is a quadratic. We are given that <math>n</math> is one of the roots. Let <math>p</math> be the other root. Notice that since <math>n+p=mj</math> we have that <math>p</math> is an integer, and so from <math>np=m+j</math> we have that <math>p</math> is positive. | |
− | Suppose <math>\frac{n^3+1}{mn-1}=k</math> where <math>k</math> is a positive integer. Then <math>n^3+1=(mn-1)k</math> and so it is clear that <math>k\equiv -1\pmod{n}</math>. So, let <math>k=jn-1</math> where <math>j</math> is a positive integer. Then we have <math>n^3+1=(mn-1)(jn-1)=mjn^2- | ||
It is obvious that <math>j=m=n=p=2</math> is a solution. Now, if not, and <math>j,m,n,p</math> are all greater than <math>1</math>, we have the inequalities <math>np>n+p</math> and <math>mj>m+j</math> which contradicts the equations <math>np=m+j, n+p=mj</math>. | It is obvious that <math>j=m=n=p=2</math> is a solution. Now, if not, and <math>j,m,n,p</math> are all greater than <math>1</math>, we have the inequalities <math>np>n+p</math> and <math>mj>m+j</math> which contradicts the equations <math>np=m+j, n+p=mj</math>. | ||
Line 16: | Line 15: | ||
==See also== | ==See also== | ||
+ | {{IMO box|year=1994|num-b=3|num-a=5}} | ||
+ | [[Category:Olympiad Number Theory Problems]] |
Latest revision as of 12:28, 26 April 2009
Problem
Find all ordered pairs where and are positive integers such that is an integer.
Solution
Suppose where is a positive integer. Then and so it is clear that . So, let where is a positive integer. Then we have which by cancelling out the s and dividing by yields . The equation is a quadratic. We are given that is one of the roots. Let be the other root. Notice that since we have that is an integer, and so from we have that is positive.
It is obvious that is a solution. Now, if not, and are all greater than , we have the inequalities and which contradicts the equations . Thus, at least one of is equal to .
If one of is , without loss of generality assume it is . Then we have . That is, which gives positive solutions . These give and since we assumed , we can also have and .
If one of is , without loss of generality assume it is . Then we have . That is, which gives positive solutions . These give and since we assumed , we can also have and .
From these, we have all solutions .
See also
1994 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |