sum of cubes

[bugzpodder]

we have m^3+n^3+99mn=33^3

find the number of pairs of integers (m,n) for which mn>=0

well, i see that LHS is the expansion of:

(m+n)^3=m^3+3(m+n)mn+n^3

so m+n=33 or m=0 and n=0
we have a total of 34 solutions ie (0,33),(1,32),...,(33,0)

but its a multiple choice question with solutions A)2 B) 3 C)33 D)35 E)99

its driving me crazy that my answer is between 33 and 35!! what am i missing?

[sql_lall]

you have:
(m+n)^3=m^3+3(m+n)mn+n^3

and say "if m+n = 33, then the equation is true"

however, can you prove it is if AND ONLY IF

[krtxmrtz]

Originally posted by bugzpodder
its a multiple choice question with solutions A)2 B) 3 C)33 D)35 E)99

its driving me crazy that my answer is between 33 and 35!!

Do you mean you know that the correct answer is not 33, 34 or 35? At first glance I'd say 35...

[krtxmrtz]

Forget my previous post: I had counted 0,0 whcih can't be counted in. So I guess 34.

[bugzpodder]

m=n=-33 also works. i overlooked mn>=0 and thought they are all positive. so its 35