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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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
I already found, satisfy this equation. – 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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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