I have tried to get representations of some integers as sum of three cubic of the form $x^3+(k\*10^n)^3+z^3$ with $k$ is integer and $n$ is a postive integer, I took this example : $(48807585839879)^3-(4\times10^{14})^3+z^3=0$, I have got $z=399757627176339$ look [here](https://www.wolframalpha.com/input/?i=solve+in+integers+%2848807585839879%29%5E3-%284*10%5E14%29%5E3%2Bz%5E3%3D0), but in the fact this is not a solution after substitution in precedent equation and this contradict the first Fermat theorem which claim that there is no integers $x, y, z$ with $xyz\neq 0$ satisfy :$ x^3 + y^3 + z^3=0 $ , what is the numerical interpretation of that solution ? I have got many examples by taking $y=k\*10^n$ really this mixed me so much



Why wolfram alpha gives integers solutions for some equations of the form $ x^3 +(k\times10^n)^3 + z^3=0 $?
Not that I'm a great fan of Mathematica, but Wolfram Alpha yields the same thing as Will Jagy got with PARI:


* Input: 48807585839879^3 + 399757627176339^3
* Result: 63999999999999992173445324722427124758794658


I think that the issue is that the default precision in Mathematica is not set high enough to handle these numbers when doing real calcualtions. In PARI with precision set to 38 decimal digits, the real root of


48807585839879^3 - (4\*10^14)^3+z^3


is 399757627176339.01632510004931888021305, so one can see it's not an integer. On the other hand, if I set the real precision of PARI to be 16 digits, then PARI returns the real root 399757627176339.0, so one could possibly be fooled in PARI, too. So the lesson isn't so much "don't use Alpha" as it is "be sure you're computing to sufficient precision for the numbers involved." For example, the following command in Alpha will yield a result that clearly indicates that the root is not an integer:


Solve 48807585839879^3 - (4\*10^14)^3+z^3=0 for z with precision 30 digits


