# Thread: Prove that a^3 - b^3 = 2011 has no integer solutions.

1. ## Prove that a^3 - b^3 = 2011 has no integer solutions.

I am a number theory newbie. So please explain all the steps.

2. Hello,
Originally Posted by fardeen_gen
I am a number theory newbie. So please explain all the steps.
You can see that 2011 is a prime number. Thus its only possible factorisation is $\pm$ 1 $\cdot$ $\pm$ 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.
Therefore a 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

$\Delta$=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 ^^

3. Hello, fardeen_gen!

A slight variation of Moo's solution . . .

Prove that .a³ - b³ .= .2011 .has no integer solutions.

We have: .(a - b)(a² + ab + b²) .= .2011

Since 2011 is prime, the factors are .±1 and ±2011

We have: .a - b .= .±1 . . a .= .b ± 1 . [1]

. . . .and: .a² + ab + b² .= .±2011 . [2]

Substitute [1] into [2]: .(b ± 1)² + b(b ± 1) + b² .= .±2011

This simplifies to two equations:

. . . . .b² + b - 670 .= .0
. . . . . . . . . . . . . . . . . . ... and neither has integer solutions.
. . 3b² - 3b + 2012 .= .0

4. Correcting some omissions:

5. A proof by using congruences