# Name of Formula $x^3+y^3=z^3+1$

I encountered the formula $$x^3+y^3=z^3+1$$ with the condition, that $$x<y<z$$ and wonder, whether it has got a specific name or whether it can be easily transformed into another well-known (family of) formula(s).

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Hi 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. –  Glen Wheeler Apr 24 '11 at 20:10
In 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 $x$. –  Torbjoern Apr 25 '11 at 10:19
@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] (en.wikipedia.org/wiki/Diophantine_equation) –  night owl Apr 26 '11 at 7:20
It looks like the equation for a one-sheet hyperboloid. –  please delete me Apr 26 '11 at 7:23
@nightowl: I think, the term Diphantine equations is the thing I was looking for. Thank you very much for this answer/comment. –  Torbjoern Apr 26 '11 at 15:35

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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Thank you for this link and for the formula. All elements of the set $\{(x,y,z), x<y<z:x3+y3=z3+1\}$ I already found, satisfy this equation. –  Torbjoern Apr 26 '11 at 15:40

$$X^3+ Y^3+ Z^3=1$$ 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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Maybe you want to give any reference to this? Searching for "harder factor" doesn't yield anything interesting on google & co. –  Listing Apr 24 '11 at 10:28
I don't see how this is related to the original question. Could you please elaborate? –  t.b. Apr 24 '11 at 13:16
This is very wrong. Among other issues, $X$,$Y$, and $Z$ are not integers! –  Glen Wheeler Apr 24 '11 at 19:58
@GlenWheeler: He's right, that $X$, $Y$ and $Z$ are integers --- at least in the formula I encountered. –  Torbjoern Apr 25 '11 at 10:22
@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. –  Glen Wheeler Apr 25 '11 at 10:52