Difference between revisions of "2023 AMC 10B Problems/Problem 14"
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Putting all cases together, the total number of solutions is | Putting all cases together, the total number of solutions is | ||
− | \boxed{\textbf{(C) 3}}. | + | <math>\boxed{\textbf{(C) 3}}</math>. |
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
Revision as of 20:02, 15 November 2023
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)