Problem:
The number of ordered pairs of integers (m,n) for which mn≥0 and
m3+n3+99mn=333
is equal to
Answer Choices:
A. 2
B. 3
C. 33
D. 35
E. 99
Solution:
Let m+n=s. Then m3+n3+3mn(m+n)=s3. Subtracting the given equation from the latter yields
s3−333=3mns−99mn
It follows that (s−33)(s2+33s+332−3mn)=0, hence either s=33 or (m+n)2+33(m+n)+332−3mn=0. The second equation is equivalent to (m−n)2+(m+33)2+(n+33)2=0, whose only solution, (−33,−33), qualifies. On the other hand, the solutions to m+n=33 satisfying the required conditions are (0,33),(1,32),(2,31),…,(33,0), of which there are 34. Thus there are 35 solutions altogether.