5
$\begingroup$

Well, here's my question:

Are there any integers, $a$ and $b$ that satisfy the equation $b^2$$+4$=$a^3$, such that $a$ and $b$ are coprime?

I've already found the case where $b=11$ and $a =5$, but other than that? And if there do exist other cases, how would I find them? And if not how would I prove so?

Thanks in advance. :)

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    Well, then you should also remove the part asking for the proof of their non-existence-if-they-don't in your edit. Tricky among other things.2012-12-22
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    Oh, yikes. I'm sorry. You're right. It was misleading. I think it's okay now, though.2012-12-22

4 Answers 4

5

Update: This is a Mordell equation and from the ref E_-00004 from this table all the known solutions were provided here :

E_-00004: r = 1   t = 1   #III =  1
          E(Q) = <(2, 2)>
          R =   0.4503206856
           4 integral points
            1. (2, 2) = 1 * (2, 2)
            2. (2, -2) = -(2, 2)
            3. (5, 11) = -2 * (2, 2)
            4. (5, -11) = -(5, 11)

Fine references about this kind of problems are :

  • de Jonquières' 1878 paper (french)
  • Conrad's paper for simple impossibilities proofs but not only since the theorem $3.3$ is the proof that no other solutions in $\mathbb{Z}$ exists for your equation.

In Jonquières' paper one finds "D'autres fois, mais rarement, on démontre qu'il n'existe qu'une seule solution. C'est ce qui a été fait par Fermat, Euler et Legendre pour les équations $x^3-2=y^2$, $x^3-4=y^2$...¨.

This means that no other solution exist and that this was proved by one or more between Fermat, Euler and Legendre (I'll search references).

4

$a=5, b=11$ is one satisfying it. I don't think this is the only pair.

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    You didn't read the question, did you?2012-12-22
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    Oh, i very well did. but you seem to have missed the first version of his Q. he added the line starting "Ah well.. " after my answer appeared here-- see my comment up there. he is getting a flag. he is a cheat.2012-12-22
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    I think it's important not to write offensive or even "tough" commentaries towards other people since, unfortunately, it is a *very common* practice over here to change the original question, either by the OP or by someone else. Ashley's answer correctly addressed the OP. +12012-12-22
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    I apologize. I wrongly interpreted the time of your answer. It's odd that a question can be edited for a while without leaving any mark.2012-12-22
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    Oh, I never added the line mentioning 5 and 11. That was always a part of the original question. But I concede, my original question was misleadling-ly phrased.2012-12-22
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    @DonAntonio i see your point on the offensive language and do agree to it. i also think that it is an unpleasantness to the user of the language before it is to many others. however, i despise more seeing such petty "efforts" finding grounds on the site-- and this time at the cost of showing me as the one in the wrong. plagiarism is plagiarism. it never is pleasant to deal with but it never should leak in. thank you for your comment.2012-12-22
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    Karolis Juodelė no problem. i probably would have thought the same. thanks for your comment.2012-12-22
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    @MWarsi: you must be meaning you never added that line in your __original__ Q. this is my last comment to you on this Q and very probably anywhere else.2012-12-22
3

Using Gaussian integers it is easy to show that the general solution of $$x^2+y^2=z^3$$ is $$x=m^3-3m n^2$$ $$y=3m^2n-n^3$$ $$z=m^2+n^2$$

If $x=2$ you get $m=\pm 1, \pm2$ and then you can solve the problem.

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    Hey, thanks for that. Now, I've never seen it done that way though. How did you derive the expressions for x, y and z?2012-12-23
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    You factor $x^2+y^2=(x+iy)(x-iy)$. The two factors are relatively prime in $Z[i]$. Therefore $$x+iy=(m+ni)^3$$.2012-12-23
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    The above derivation is incorrect, for example if $x=y=2$ then $x+iy$ and $x-iy$ are not relatively prime.2014-07-21
0

Observe that

if $a=3k,b^2=a^3-4=(3k)^3-4\equiv-1\pmod 3$ but $b^2\equiv1,0\pmod 3$

if $a=3k+1,b^2=a^3-4=(3k+1)^3-4=9(3k^3+3k^2+k)-3$ which is divisible by $3,$ but not by $9$

So, $a$ must be of the from $3k+2$

Consequently, $b^2-4=a^3-8=(3k+2)^3-8$

$(b+2)(b-2)=9k(3k^2+6k+4)$

Also, as $(a,b)=1,$ both $a,b$ must be odd $\implies (b+2,b-2)=(b+2,b+2-(b-2))=1$ and $k$ is odd

As $k$ is odd, $(k,3k^2+6k+4)=(k,4)=1$

If $b-2=9k,b+2=9k+4$ and $b+2=3k^2+6k+4\implies 3k^2-3k=0\implies k=0,1$

$k=0\implies b=2,a=2$ (but both $a,b$ are odd)

$k=1\implies b=11,a^3=125,a=5$

If $b+2=9k,b-2=9k-4$ and $b-2=3k^2+6k+4\implies 3k^2-3k+8=0$ whose discriminant is negative.

If $b+2=9,a^3=b^2+4=53$

If $b-2=9\implies b=11,a^3=b^2+4=125\implies a=5$

If $b-2=k\implies b=k+2,b+2=k+4$ and $b+2=9(3k^2+6k+4),27k^2+53k+32=0$ whose discriminant is negative.

If $b+2=k\implies b=k-2,b-2=k-4$ and $b-2=9(3k^2+6k+4),27k^2+53k+40=0$ whose discriminant is negative.

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    But what if each factor on the left-hand side shares a factor with two or more factors on the right-hand side? I don't believe your proof considers that possibility.2014-11-13