# A number theory question that is probably wrong

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

I think the question is wrong as

$$a^3-b^3 = 2011$$

$$(a-b)(a^2+ab+b^2) = 2011$$

As $$2011$$ is prime so the only factors $$2011$$ has are $$1$$ and $$2011$$ itself.

So if $$a-b = 1$$ and $$a^2+ab+b^2 = 2011$$ , then we can say that

$$a^2-2ab+b^2 = 1$$

So , $$a^2+b^2 = 2ab+1$$

Putting the value in the second equation

$$3ab +1 = 2011$$

$$3ab = 2010$$

And as $$2010$$ is divisible by $$3$$ it has integer solutions , which is contradictory to the original question. Can anybody help me with this?

• Questions are made by human beings, so they are wrong sometimes. What values of $a$ and $b$ did you find? Did you check they satisfy $a^3-b^3=2011$? Note that you are looking for factors $a$ and $b$ of $2010/3=670$ such that $a-b=1$. Do such numbers exist? May 24 at 10:53
• $25 \times 26 < 670 < 26 \times 27.$ May 24 at 10:56
• You need to verify that the divisors of $ab=670$ with $a-b=1$ really satisfy $a^3-b^3=2011$. So far there is no contradiction to the original question as you claim. May 24 at 10:57
• @user2661923 Exactly, this is what I said. Everything lines up with the original question. There is no mistake as claimed. Indeed, there really are no solutions. May 24 at 10:59
• Yes, $3ab = 2010$ has integer solutions, but remember you also supposed $a-b=1$, which would need to be true at the same time. May 24 at 11:06

Well, you seem to be doing a very good job up until when you realize that

one must have $$(a-b)(a^2+ab+b^2)= 2011$$.

You then come to the key observation $$a-b$$ must be a divisor of $$2011$$.

At this point I would consider the problem "almost solved", it seems like now we must only do some case checking and everything will be fine.

Case $$1$$: $$a-b = 1$$. We get $$b=a-1$$ and so we have $$a^2 + (a-1)a + a^2 = 2011$$.

Case $$2$$: $$a-b= 2011$$. We get $$b= a-2011$$ and so we must have $$(2011)(a^2 + (a-2011)a + (a-2011)^2) = 2011$$.

Case $$3$$: $$a-b= -1$$. We get $$b= a+1$$ and so we must have $$-(a^2 + (a+1)a + (a+1)^2) = 2011$$.

Case $$4$$: $$a-b= -2011$$. We get $$b= a+2011$$ and so we must have $$-2011(a^2 + (a+2011)a + (a+2011)^2) = 2011$$.

Now, do these cases have solutions? These are the sort of equations that can be "solved" with the quadratic formula ! So you can get the answers needed.

The possible values or $$x^3\pmod 9$$ are $$\{0,1,-1\}$$ only.

So their difference can only take the values $$\{0,1,2,-1,-2\}$$, yet $$2011\equiv 4\pmod 9$$ so it is not reachable.