2016 AIME I Problems/Problem 11
Contents
Problem
Let be a nonzero polynomial such that for every real , and . Then , where and are relatively prime positive integers. Find .
Solution 1
Plug in to get . Plug in to get . Plug in to get . So for some polynomial . Using the initial equation, once again, From here, we know that for a constant ( cannot be periodic since it is a polynomial), so . We know that . Plugging those into our definition of : or . So we know that . So . Thus, the answer is .
Solution 2
From the equation we see that divides and divides so we can conclude that and divide (if we shift the function right by 1, we get , and from here we can see that divides ). This means that and are roots of . Plug in and we see that so is also a root.
Suppose we had another root that is not one of those . Notice that the equation above indicates that if is a root then and is also a root. Then we'd get an infinite amount of roots! So that is bad. So we cannot have any other roots besides those three.
That means . We can use to get . Plugging in is now trivial and we see that it is so our answer is
Solution 3
Although this may not be the most mathematically rigorous answer, we see that . Using a bit of logic, we can make a guess that has a factor of , telling us has a factor of . Similarly, we guess that has a factor of , which means has a factor of . Now, since and have so many factors that are off by one, we may surmise that when you plug into , the factors "shift over," i.e. , which goes to . This is useful because these, when divided, result in . If , then we get and , . This gives us and , and at this point we realize that there has to be some constant multiplied in front of the factors, which won't affect our fraction but will give us the correct values of and . Thus , and we utilize to find . Evaluating is then easy, and we see it equals , so the answer is
Solution 4
Substituting into the given equation, we find that . Therefore, either or . Now for integers , we know that Applying this repeatedly, we find that If , this shows that has infinitely many roots, meaning that is identically equal to zero. But this contradicts the problem statement. Therefore, , and we find for all positive integers . This cubic polynomial matches the values for infinitely many numbers, hence the two polynomials are identically equal. In particular, , and the answer is .
Solution 5
We can find zeroes of the polynomial by making the first given equation . Plugging in and gives us the zeroes and , respectively. Now we can plug in these zeros to get more zeroes. gives us the zero (no pun intended). makes the equation , which means is not necessarily . If , then plugging in to the equation yields , plugging in to the equation yields , and so on, a contradiction of "nonzero polynomial". So is not a zero. Note that plugging in to the equation does not yield any additional zeros. Thus, the only zeroes of are and , so for some nonzero constant . We can plug in and into the polynomial and use the second given equation to find an equation for . and , so: Plugging in into the polynomial yields . .
Solution 6 (very Easy)
Plug in yields . Since also , we have and . Plug in yields so .
Repeat the action gives , , , , and .
Since is a polynomial, the th difference is constant, where . Thus we can list out the 0th, 1st, 2nd, 3rd, ... differences until we obtain a constant.
Since the 3rd difference of is constant, we can conclude that .
Let . Plug in the values for and solve the system of 4 equations gives
Thus and
~ Nafer
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.