Difference between revisions of "2023 AMC 10B Problems/Problem 14"
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How many ordered pairs of integers <math>(m, n)</math> satisfy the equation <math>m^2+mn+n^2 = m^2n^2</math>? | How many ordered pairs of integers <math>(m, n)</math> satisfy the equation <math>m^2+mn+n^2 = m^2n^2</math>? | ||
| + | |||
| + | == Solution == | ||
| + | |||
| + | Obviously, <math>m=0,n=0</math> is a solution. | ||
| + | |||
| + | <cmath>\begin{align*} | ||
| + | m^2+mn+n^2 &= m^2n^2\\ | ||
| + | m^2+mn+n^2 +mn &= m^2n^2 +mn\\ | ||
| + | (m+n)^2 &= m^2n^2 +mn\\ | ||
| + | (m+n)^2 &= mn(mn+1)\\ | ||
| + | \end{align*}</cmath> | ||
| + | |||
| + | This basically say that the product of two consecutive numbers <math>mn,mn+1</math> must be a perfect square which is practically impossible except <math>mn=0</math> or <math>mn+1=0</math>. | ||
| + | <math>mn=0</math> gives <math>(0,0)</math>. | ||
| + | <math>mn=-1</math> gives <math>(1,-1), (-1,1)</math>. | ||
| + | |||
| + | ~Technodoggo | ||
Revision as of 16:08, 15 November 2023
How many ordered pairs of integers 
satisfy the equation 
?
Solution
Obviously, 
is a solution.
This basically say that the product of two consecutive numbers 
must be a perfect square which is practically impossible except 
or 
.
gives 
.
gives 
.
~Technodoggo
