1991 AIME Problems/Problem 1
Problem
Find if and are positive integers such that
Contents
Solution
Solution 1
Define and . Then and . Solving these two equations yields a quadratic: , which factors to . Either and or and . For the first case, it is easy to see that can be (or vice versa). In the second case, since all factors of must be , no two factors of can sum greater than , and so there are no integral solutions for . The solution is .
Solution 2
Since , this can be factored to . As and are integers, the possible sets for (ignoring cases where since it is symmetrical) are . The second equation factors to . The only set with a factor of is , and checking shows that it is correct.
Solution 3
Let , then we get the equations After finding the prime factorization of , it's easy to obtain the solution . Thus Note that if , the answer would exceed which is invalid for an AIME answer. ~ Nafer
Solution 4
From the first equation, we know . We factor the second equation as . Let and rearranging we get . We have two cases: (1) and OR (2) and . We find the former is true for . .
See also
1991 AIME (Problems • Answer Key • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
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