2007 AMC 10A Problems/Problem 23
Problem
How many ordered pairs 
of positive integers, with 
, have the property that their squares differ by 
?
Solution 1
![\[m^2 - n^2 = (m+n)(m-n) = 96 = 2^{5} \cdot 3\]](http://latex.artofproblemsolving.com/b/9/d/b9dfb5d5b4fe15db2a54e11cc6df787e8cd77f3f.png)
For every two factors 
, we have 
. It follows that the number of ordered pairs is given by the number of ordered pairs 
. There are 
factors of , which give us six pairs 
. However, since 
are positive integers, we also need that 
are positive integers, so 
and 
must have the same parity. Thus we exclude the factors 
, and we are left with 
pairs 
.
Solution 2
Similarly to the solution above, write as 
. To find the number of distinct factors, add 
to both exponents and multiply, which gives us 
factors. Divide by 
since 
must be greater than or equal to 
. We don't need to worry about and being equal because is not a perfect square. Finally, subtract the two cases above for the same reason to get .
Solution 3
Find all of the factor pairs of : 
You can eliminate 
and (
because you cannot have two numbers add to be an even number and have an odd difference at the same time without them being a decimal. You only have pairs left, so the answer is 
.
~HelloWorld21
See also
| 2007 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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