2023 AMC 10B Problems/Problem 14
How many ordered pairs of integers satisfy the equation
?
Solution 1
Clearly, is 1 solution. However there are definitely more, so we apply Simon's Favorite Factoring Expression to get this:
This basically say that the product of two consecutive numbers must be a perfect square which is practically impossible except
or
.
gives
.
gives
.
~Technodoggo ~minor edits by lucaswujc
Solution 2
Case 1: .
In this case, .
Case 2: .
Denote .
Denote
and
.
Thus,
.
Thus, the equation given in this problem can be written as
Modulo , we have
.
Because
, we must have
.
Plugging this into the above equation, we get
.
Thus, we must have
and
.
Thus, there are two solutions in this case: and
.
Putting all cases together, the total number of solutions is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3 (Discriminant)
We can move all terms to one side and wrote the equation as a quadratic in terms of to get
The discriminant of this quadratic is
For
to be an integer, we must have
be a perfect square. Thus, either
is a perfect square or
and
. The first case gives
, which result in the equations
and
, for a total of two pairs:
and
. The second case gives the equation
, so it's only pair is
. In total, the total number of solutions is
.
~A_MatheMagician