I encountered the formula
$$x^3+y^3=z^3+1$$
with the condition, that
$$x
Name of Formula $x^3+y^3=z^3+1$
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0Hi T.K.. How did you come upon this formula? The set of $(x,y,z)$ which satisfy your conditions is some subset of $\mathbb{R}^3$. Your condition is cone-like, so the set $\{(x,y,z):x^3+y^3 = z^3 + 1\}$ is contained in a kind of cone. – 2011-04-24
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0In a programming class this formula was part of the task: Write a program, that calculates the first 18 sets of $(x,y,z)\in\mathbb{R}^+$, which satisfy the formula and the condition, ordered by increasing $x$. – 2011-04-25
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1@T.K.: I am not sure because I have yet to encounter equations personally yet, but I do run across things from time to time and this looked familiar when I seen it. I am thinking that this could be an example of a *Diophantine equations*, but like I said not sure so don't take that as a for sure. Might want to give this a look. [1] (http://en.wikipedia.org/wiki/Diophantine_equation) – 2011-04-26
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0It looks like the equation for a one-sheet hyperboloid. – 2011-04-26
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0@nightowl: I think, the term _Diphantine equations_ is the thing I was looking for. Thank you very much for this answer/comment. – 2011-04-26
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0@T.K.: Sure, your welcome.. Anytime. – 2011-04-28
2 Answers
Have a look at http://www.mathpages.com/home/kmath071.htm
There you will find $$(1\pm9m^3)^3+(9m^4)^3+(-9m^4\pm3m)^3=1$$ and another similar-but-more-complicated formula, also it says it is known that there are infinitely many such formulas, and it is not known whether every solution is part of such an infinite family.
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0Thank you for this link and for the formula. All elements of the set $\{(x,y,z), x
– 2011-04-26
$$X^3+ Y^3+ Z^3=1$$ is the formula which is known as harder factor and yours is a distorted and conditional form of harder factor
If $X+Y+Z=0$ then $X^3+ Y^3+ Z^3=1$.
In your question $X$ is less than $Y$ and $Y$ is less than $Z$ means the minimum possible difference between $X$ and $Y$, $Y$ and $Z$ is $1$. At the same time the minimum possible difference between $X$ and $Z$ will be $2$.
So there will in all the cases except $X=-2$, $Y=-1$, $Z=3$ where $X+Y+Z$ is not equal to zero then it must be that $X^3+ Y^3+ Z^3$ is not equal to $1$. So $X^3+ Y^3+ Z^3$ must be greater/less than $1$. As it is given that $Z>Y>X$ then $X^3+ Y^3$ must be unequal to $Z^3$. It means $X^3+ Y^3$ may be equal to $Z^3+1$.
In this way $X^3+ Y^3+ Z^3=1$ is related to the question asked by the poster
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3Maybe you want to give any reference to this? Searching for "harder factor" doesn't yield anything interesting on google & co. – 2011-04-24
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0I don't see how this is related to the original question. Could you please elaborate? – 2011-04-24
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1This is very wrong. Among other issues, $X$,$Y$, and $Z$ are *not* integers! – 2011-04-24
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0@GlenWheeler: He's right, that $X$, $Y$ and $Z$ _are_ integers --- at least in the formula I encountered. – 2011-04-25
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0@T.K. Ok, but then you should mention this in the question *and* should not use $\mathbb{R}$. This is the set of real numbers. – 2011-04-25