Hello,

You can see that 2011 is a prime number. Thus its only possible factorisation is 1 2011. Why do I say that ? It's because we'll factorise a^3-b^3.

a^3-b^3=(a-b)(a^2+ab+b^2)

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Let's see what the signs of a & b are.

If a<0, then a^3<0 and so -b^3>0 that is to say b<0. And b^3>a^3 --> b>a --> b-a>0, thus a^2+ab+b^2<0, which is not possible because a²+b²>0 and a & b have the same sign and then ab>0.

Thereforea has to be positive.

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Let's assume that a & b > 0.

Then a²+ab+b² > 1. Thus the only possibility is that a²+ab+b²=2011 and a-b=1.

--> a=b+1. Substituting in a²+ab+b²=2011 :

b²+2b+1+b²+b+b²=2011.

3b²+3b=2010

b²+b=670

b²+b-670=0

=1+4*670=2681. But 2681 is not a perfect square. Thus b is not an integer. Which is not satisfying the conditions. Therefore, there are no a & b, positive integers, such that a^3-b^3=2011.

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What if a>0 and b<0 ?

Well, try to think about that

@CB : not sufficient, eh ?

Edit : my 27th ^^