2012 AMC 10A Problems/Problem 24
Let , , and be positive integers with such that and .
What is ?
Add the two equations.
Now, this can be rearranged:
, , and are all integers, so the three terms on the left side of the equation must all be perfect squares. Recognize that .
, since is the biggest difference. It is impossible to determine by inspection whether or , or whether or .
We want to solve for , so take the two cases and solve them each for an expression in terms of . Our two cases are or . Plug these values into one of the original equations to see if we can get an integer for .
, after some algebra, simplifies to . 2021 is not divisible by 7, so is not an integer.
The other case gives , which simplifies to . Thus, and the answer is .