2012 AMC 10A Problems/Problem 24
Problem 24
Let ,
, and
be positive integers with
such that
and
.
What is ?
Solution
Add the two equations.
.
Now, this can be rearranged:
and factored:
,
, 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
.