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Does anyone know how to get all solutions of the equation $a^3=b^3+c^3$ when $b,c$ quadratic irrationals or $b$ rational and $c$ quadratic irrational?

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To emphasize the importance of answering jspecter's question, take arbitrary $b$ and $c$ satisfying the given conditions, then define $a=\sqrt[3]{b^3+c^3}$ and you have described all solutions to the problem as stated. –  Jonas Meyer Dec 24 '11 at 4:38
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@Vassili: That would be better edited into the question. –  Jonas Meyer Dec 24 '11 at 4:58
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@QED: Off topic, but I am one of the 2 people who has downvoted this question "so heavily" and I did it because of this post, not because of any other post by the same user, and certainly not because of the person. –  Jonas Meyer Dec 24 '11 at 18:52
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@QED: I have just downvoted; the post fits the first reason indicated when you mouse over the downvoting arrow: "This question does not show any research effort." If Vassili does manage to explain why he's looking at these things in the first place, and edits the explanation into this question, I'll be more than willing to revoke my downvote. –  J. M. Dec 25 '11 at 4:36
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I have been repeating repeating and repeating, trust me, unless you select an answer for your question promptly and unless you improve your formatting you are not going to get up-voted, don't think me as if I am suggesting you and commanding you, I am not in a position to do so, but previously I have experienced the same things, but now I changed my self so community also changed the way they behaved with me, as world is like mirror, it appears the way you look at it @Vassili –  Iyengar Dec 25 '11 at 5:20

2 Answers 2

up vote 3 down vote accepted

Here is an explicit calculation of what jspecter outlined:

The line described is $c = ma - m$ and with this $a^3 - c^3 - 1 = 0$ reduces to $(-m^3 + 1)a^3 + 3m^3a^2 - 3m^3a + (m^3 - 1) = 0$. Dividing both sides by $a-1$ (to factor out the point $O$) leaves us with $(-m^3 + 1)a^2 + (2m^3 + 1)a + (-m^3 + 1)$ which has roots $$a = \frac{1 + 2m^3 \pm \sqrt{12m^3 - 3}}{2 - 2m^3}.$$

For example if we let $m = 7$ we find the pair $(a,c) = (\frac{229 + \sqrt{457}}{228},\frac{7 + 7 \sqrt{457}}{228})$ and you can check that $a^3 = c^3 + 1$. In particular $$(7+7\sqrt{457})^3 + 228^3 = (229+\sqrt{457})^3.$$

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:The number 228 is an integer not an irrational number. –  Vassilis Parassidis Dec 25 '11 at 5:34
    
@Vassili, why do you tell me this? –  user16697 Dec 25 '11 at 5:36
    
:because i ask for all solutions. –  Vassilis Parassidis Dec 25 '11 at 5:41
    
@Vassili, I don't understand. –  user16697 Dec 25 '11 at 5:41
    
I want to know if with the above method we obtain all solutions. All the solutions of the sum of two cubes with the outlined conditions contain a case where all three numbers are quadratic irrationals. –  Vassilis Parassidis Dec 25 '11 at 5:48

Here's an idea to think about.

Assume $b \ne 0.$ In this case, it is equivalent to classify all pairs $(A,C)$ where $A$ and $C$ lie a quadratic extension of $\mathbb{Q}$ and satisfy the equation

$$A^3 - C^3 = 1. \mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ }\mbox{ } (1)$$

We denote by $V$ the set of all pairs in $\overline{\mathbb{Q}}$ which satisfy $(1).$

Note $O := (1,0)$ is such a pair. Fix a slope $m \in \mathbb{Q}$ (you can also take $m = \infty$) and consider line $L$ of slope $m$ through $O.$ The line $L$ intersects $V$ in three points and each coordinate of these points is cut out by a cubic polynomial in one varriable over $\mathbb{Q}.$ However, one of the points of intersection is $O,$ so one root of each of the polynomials is rational. Factoring out this root, it follows that the remaining roots satisfy a degree 2 rational polynomial and therefore lie in quadratic extensions of $\mathbb{Q}.$

Now this won't get you all the points, for example $P = (0,\zeta_6)$ lies on $V,$ but doesn't lie on a line through $O$ with rational slope. However, $P$ does lie on the line through $(0,-1) \in V$ with infinite slope. So the above technique can be adapted to obtain $P.$

I claim that the idea above can be used to obtain all points on $V$ in quadratic extensions. That is if you know all points on $V$ in $\mathbb{Q}.$

But even Fermat "knew" that.

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BTW, a good resource to look at would be the first couple pages of Rational Points on Elliptic Curves by Silverman and Tate. amazon.com/Rational-Points-Elliptic-Undergraduate-Mathematics/… –  jspecter Dec 24 '11 at 5:29
    
@jspecter.Can you give me an examble where a,b,c are all quadratic irrationals. –  Vassilis Parassidis Dec 24 '11 at 5:58
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Try to follow through the recipe with different slopes. See what you get. I would post one, but I have to catch a flight. –  jspecter Dec 24 '11 at 6:07
    
can someone give me an example on posted question where a,b,c are all quadratic irrationals? –  Vassilis Parassidis Dec 24 '11 at 6:42
    
@Gerry, Vassili, I've cleaned up some of the more chatty comments on this post. –  Willie Wong Feb 24 '12 at 13:51

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