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I want to know if the equation $x^3+y^3+z^3+t^3=10^{2021}$ has distinct positive integer solutions PowersRepresentations[10^2021, 4, 3]
return

PowersRepresentations::ovfl: Overflow occurred in computation.

FindInstance[{x^3 + y^3 + z^3 + t^3 == 10^2021, 0 < x < y < z < t}, {x,y,z,t}, Integers]

My computer runs too long. How can I reduce timing to solve this equation?

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    $\begingroup$ This looks like an exercise made to be impossible to solve by brute force. You probably have to consider the equation modulo a couple of suitably chosen prime numbers or use some other number theoretical trick. $\endgroup$ Commented Feb 17, 2021 at 10:46
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    $\begingroup$ Ask it at MSE which is a right forum for such questions. $\endgroup$
    – user64494
    Commented Feb 17, 2021 at 10:49
  • $\begingroup$ E.g. see that article. $\endgroup$
    – user64494
    Commented Feb 17, 2021 at 11:56
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    $\begingroup$ This is related to Waring's problem and it is conjectured that 7,373,170,279,850 is the largest integer which cannot be expressed as the sum of four nonnegative integral cubes. This is different from proving that such representation exists for $10^{2021}$. $\endgroup$
    – kirma
    Commented Feb 17, 2021 at 13:01
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    $\begingroup$ As a general rule of thumb, if a problem has a year in the last 50 years as a parameter in it, it was created for some kind of contest purpose (in that year) where people would be working by hand. This is already a useful principle because it suggests that something is going on that ought to make it manually solvable. $\endgroup$ Commented Feb 18, 2021 at 4:36

2 Answers 2

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Easy, notice that $10^{2021}=100\times 10^{3\times 673}$. Next use your code, but for the factor 100.

FindInstance[{x^3 + y^3 + z^3 + t^3 == 100, 0<x<y<z<t}, {x,y,z,t}, Integers]

yielding a single result

(*{{x -> 1, y -> 2, z -> 3, t -> 4}}*)

Now verify the solution

(x^3 + y^3 + z^3 + t^3 /. {x -> 1 10^673,y -> 2 10^673,z -> 3 10^673,t -> 4 10^673}) == 10^2021
(* True*)
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    $\begingroup$ Since 100 is the square of the 4th triangular number, it's also the sum of the first four cubes, so someone with sufficient insight can solve this without having to calculate anything. $\endgroup$
    – Neil
    Commented Feb 18, 2021 at 0:42
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Two solutions may be found starting from: $$10^{2021}=10^5\times10^{2016}=12500\times2^3\times10^{3\times672}$$

Since $12500=19^3+17^3+8^3+6^3=18^3+17^3+12^3+3^3$ we have:

$$10^{2021}=(38\times10^{672})^3+(34\times10^{672})^3+(16\times10^{672})^3+(12\times10^{672})^3$$ $$10^{2021}=(36\times10^{672})^3+(34\times10^{672})^3+(24\times10^{672})^3+(6\times10^{672})^3$$

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