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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, 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

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    $\begingroup$ if this just teaches you not to trust computers, some progress has been made. Meanwhile. I took your $x,z$ and asked pari-gp for $x^3 + z^3,$ which came out 63999999999999992173445324722427124758794658 which has small relative error from $64 \cdot 10^{42}$ $\endgroup$
    – Will Jagy
    Apr 18, 2021 at 23:46
  • $\begingroup$ @WillJagy,Thank you , you are really right because the titled equation made my day when i plugged 114 in the RHS of the titled equaton i got the same integer solution such that i tought that i have solved the problem ,I should avoid using wolfram alpha in the futur $\endgroup$ Apr 18, 2021 at 23:59
  • $\begingroup$ I think MathOverflow is rather not the right place to report software errors. -- It is likely more helpful if you report these to the developers of the respective software than here. Then the developers become aware of them, and have a chance to fix them. $\endgroup$
    – Stefan Kohl
    Apr 19, 2021 at 10:30

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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

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    $\begingroup$ Please tell me the first sentence has a typo, and you meant Mathematica and not Mathematics! $\endgroup$
    – Wojowu
    Apr 19, 2021 at 0:29
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    $\begingroup$ @Wojowu Oops. :) And thank you, Will, for correcting it! And just to clear the air, I am a huge fan of Mathematics!! $\endgroup$ Apr 19, 2021 at 10:18

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