1999 AIME Problems/Problem 3
Problem
Find the sum of all positive integers for which
is a perfect square.
nosaj's Solution
We have , where
is a positive integer. Rearranging gives us
Applying the quadratic formula yields
Now, we simplyfy:
In order for
to be an integer, the discriminant
must be a perfect square. In other words,
where
is a positive integer. Rearranging gives us
Aha! Difference of squares.
Remember that both of these factors are integers, so we have a very limited amount of choices. Either
or
These two systems yield
and
. Plugging
back in gives us
,
,
,
. Therefore, our answer is
Solution
If for some positive integer
, then rearranging we get
. Now from the quadratic formula,
Because is an integer, this means
for some nonnegative integer
. Rearranging gives
. Thus
or
, giving
or
. This gives
or
, and the sum is
.
Alternate Solution
Suppose there is some such that
. Completing the square, we have that
, that is,
. Multiplying both sides by 4 and rearranging, we see that
. Thus,
. We then proceed as we did in the previous solution.
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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