Problem:
How many ordered pairs ( m,n ) of positive integers are solutions to m4β+n2β=1?
Answer Choices:
A. 1
B. 2
C. 3
D. 4
E. more than 4
Solution:
Since m and n must both be positive, it follows that n>2 and m>4. Because m4β+n2β=1 is equivalent to (mβ4)(nβ2)=8, we need only find all ways of writing 8 as a product of positive integers. The four ways, 1β
8,2β
4,4β
2 and 8β
1, correspond to the four solutions (m,n)=(5,10),(6,6), (8,4) and (12,3).