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by harsh.champ » Thu Feb 04, 2010 6:16 am
Let x and y be consecutive integers. Suppose m be the number of solutions of x3 − y3 = 3k2 and n be the number of solutions of x3 − y3 = 2t2, where k, t are integers. Then m + n equals:

(1)31
(2)13
(3)5
(4)6
(5)None of the above
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by ajith » Thu Feb 04, 2010 7:11 am
harsh.champ wrote:Let x and y be consecutive integers. Suppose m be the number of solutions of x3 − y3 = 3k2 and n be the number of solutions of x3 − y3 = 2t2, where k, t are integers. Then m + n equals:

(1)31
(2)13
(3)5
(4)6
(5)None of the above
(x-y)(x^2+y^2+xy) = 3k2

x^2+y^2+xy = 3k2


(9^2) + (8^2) + (9 * 8) = 217

(10^2) + (9^2) + (10 * 9) = 271

(11^2) + (10^2) + (10 * 11) = 331

(11^2) + (12^2) + (12 * 11) = 397

So it is evident that there is no solution for both the equations m = 0 n=0

I go for E
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by Mom4MBA » Thu Feb 04, 2010 10:15 am
Didn't get the question nor the answer...............anybody else with the explanation.
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