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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?

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    $\begingroup$ 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? $\endgroup$
    – Taladris
    May 24 at 10:53
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    $\begingroup$ $25 \times 26 < 670 < 26 \times 27.$ $\endgroup$ May 24 at 10:56
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    $\begingroup$ 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. $\endgroup$ May 24 at 10:57
  • $\begingroup$ @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. $\endgroup$ May 24 at 10:59
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    $\begingroup$ Yes, $3ab = 2010$ has integer solutions, but remember you also supposed $a-b=1$, which would need to be true at the same time. $\endgroup$
    – aschepler
    May 24 at 11:06

2 Answers 2

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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.

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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.

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