2016 USAMO Problems/Problem 4
Problem
Find all functions such that for all real numbers and ,
Solution 1
Step 1: Set to obtain
Step 2: Set to obtain
In particular, if then
In addition, replacing , it follows that for all
Step 3: Set to obtain
In particular, replacing , it follows that for all
Step 4: Set to obtain
In particular, if , then by the observation from Step 3, because Hence, the above equation implies that , where the last step follows from the first observation from Step 2.
Therefore, either or for each
Looking back on the equation from Step 3, it follows that for any nonzero Therefore, replacing in this equation, it follows that
Step 5: If , then
This follows by choosing such that and Then , so plugging into the given equation, we deduce that Therefore, by the third observation from Step 4, we obtain , as desired.
Step 6: If , then
Suppose by way of contradiction that there exists an nonzero with Choose such that and The following three facts are crucial:
1. This is because , so by Step 5, , impossible.
2. This is because , so by Step 5 and the observation from Step 3, , impossible.
3. This is because by the second observation from Step 2, Then because , Step 5 together with the observation from Step 3 yield , impossible.
By the second observation from Step 4, these three facts imply that and and By plugging into the given equation, it follows that But the above expression miraculously factors into ! This is clearly a contradiction, since by assumption. This completes Step 6.
Step 7: By Step 6 and the second observation from Step 4, the only possible solutions are and for all It's easy to check that both of these work, so we're done.
Alternative Solution 1
From steps 1 and 2 of Solution 1 we have that , and . Therefore, if , then . Furthermore, setting gives us . The LHS can be factored as . In particular, if , then we have . However, since we have from step 2 that , assuming , the equation becomes , so for every , is equivalent to either or . From step 6 of Solution 1, we can prove that , and are the only possible solutions.
Solution 2
Step 1:
Step 2: . Now, assume . Then, if , we substitute in to get , or . Otherwise, we divide both sides by to get . If , we obviously have . Thus, the function is even. . Step 3: . Thus, , we have or .
Step 4: We now assume , . We have . Now, setting , we have or . The former implies that or . The latter implies that or . Assume the latter. . Clearly, this implies that is negative for some . Now, we have , which is a contradiction. Thus, or .
Step 5: We now assume , for some . Let be sufficiently large integer, let and take the absolute value of (since the function is even). Choose such that . Note that we have ~ and ~. Note that . Now, LHS is positive, as the second term is positive and the first term is nonnegative and thus the right term is equal to ~. Now if , the second term of the LHS/RHS clearly ~0 as . if , then we have LHS/RHS ~ , otherwise, we have LHS/RHS~~, a contradiction, as we're clearly not dividing by , and we should have LHS/RHS=1.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See also
2016 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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All USAMO Problems and Solutions |
2016 USAJMO (Problems • Resources) | ||
Preceded by Problem 5 |
Last Problem | |
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