1976 IMO Problems/Problem 2
Problem
Let and for Prove that for any positive integer n the roots of the equation are all real and distinct.
Solution
I shall prove by induction that has distinct real solutions, where are positive and are negative. Also, for ever root , .
Clearly, has 2 real solutions, where 1 is positive and 1 is negative. The absolute values of these two solutions are also both less than 2. This proves the base case.
Now assume that for some positive integer , has distinct real solutions with absolute values less than 2, where are positive and are negative.
Choose a root of . Let , where is a real root of . We have that , so , so is real and . Therefore all of the roots of are real and have absolute values less than 2.
Note that the function is an even function, since is an even function. Therefore half of the roots of are positive, and half are negative.
Now assume for the sake of contradiction that has a double root . Let . Then there exists exactly one real number such that . The only way that this could happen is when , or . However, from our inductive hypothesis, so this is a contradiction. Therefore has no double roots. This proves that that the roots of are distinct.
This completes the inductive step, which completes the inductive proof.
Alternate Solution
Let . We then have that , and we can prove using induction that . Thus, we just need to solve , which happens when or . These give that or . As we can choose the range to ensure no duplications, we get that, upon rearranging, and . There are integers in the first range and in the second. However, exactly one time, they product the same : . Thus, there are distinct roots for . We can also prove using induction that there are roots of . Thus, all roots of are distinct and real.
Q.E.D.
See also
1976 IMO (Problems) • Resources) | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |