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How can we solve $x^3 + y^3 + z^3 =57$ efficiently in a shorter way. $x$ $y$ and $z$ are integers. Given that modulus of $x$ $y$ and $z$ is less than or equal to five. We can of course do by hit and trial but what is the method of solving such questions. I actually stumbled upon this equation while solving a determinant. How to proceed. Pls help

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    $\begingroup$ Are x,y,z integers? or natural numbers? or real numbers? $\endgroup$ – Gaurang Tandon Dec 22 '17 at 16:38
  • $\begingroup$ yeah they are integers. I made the changes in the question $\endgroup$ – user354545 Dec 22 '17 at 16:46
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    $\begingroup$ Solving $x^3+y^3+z^3 = N$ in the integers is a difficult problem unless $N$ has special properties. This post may be of interest. $\endgroup$ – Tito Piezas III Dec 22 '17 at 16:53
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    $\begingroup$ Trial and error$$ \begin{array}{r|r|r} x & y & z \\ \hline -2 & 1 & 4 \\ -2 & 4 & 1 \\ 1 & -2 & 4 \\ 1 & 4 & -2 \\ 4 & -2 & 1 \\ 4 & 1 & -2 \\ \end{array}$$ $\endgroup$ – Raffaele Dec 22 '17 at 17:09
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    $\begingroup$ @TitoPiezasIII You might want to update your table as $N=74$ is solved. $$ (-284650292555885)^3 + 66229832190556^3 + 283450105697727^3 = 74 $$ $\endgroup$ – Yong Hao Ng Dec 28 '17 at 5:59
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With some forethought, you can bring it down to one case to check directly, which turns out to be a solution. A previous version of this answer examined the equation mod $7$, which led to $16$ cases to directly examine. This answer examines mod $9$, which works out even better. (The reason $7$ and $9$ are good moduli to consider is because there are relatively few cubes mod these numbers.)

Mod $9$, the only cubes are $0$, $1$, and $8$. For solutions to $X+Y+Z\equiv57\equiv3$, the only solution is $1+1+1\equiv3$. This means $x^3\equiv y^3\equiv z^3\equiv1$ mod $9$, which means $x\equiv y\equiv z\equiv1$ mod $3$. So all of $x$, $y$, and $z$ are among $\{-5,-2, 1, 4\}$.

As in @RossMillikan's answer, we can assume $x, y\leq z$ and $z$ must equal $4$. So now we must solve $x^3+y^3=-7$, with $x\leq y$ among $\{-5,-2, 1, 4\}$. Consider this equation mod $7$, where the cubes of $\{-5,-2, 1, 4\}$ are equivalent to $\{1,6, 1, 1\}$ respectively. The only way to sum two of these and get $-7\equiv0$ is using $6$ and $1$. Therefore one of $x,y$ is $-2$. Wlog, assume it's $x$, and examining $(-2)^3+y^3=-7$, we have that $y=1$ makes a complete solution $(x,y,z)=(-2,1,4)$.

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You can narrow the search with some thinking. Because of the symmetry you can demand $x \le y \le z$. Clearly $z$ has to equal $3, 4$ or $5$. If it is $5$ then $x^3+y^3=-68$, clearly $x=-4$ and there is no solution. If $z=4$ then $x^3+y^3=-7$ and $x=-2,y=1$ pops out. If $z=3, x^3+y^3=30$ and there is no solution. Not much work, helped a lot by the limit of $5$.

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    $\begingroup$ you forgot negative integers... -2, 1, 4 work. $\endgroup$ – MaxW Dec 22 '17 at 16:58
  • $\begingroup$ When $z=4$, and $x^3+y^3=-7$, you note one solution. But will there still be work to rule out any other solutions? When $z=3$, and $x^3+y^3=30$, it is still some amount of checking to establish there are no solutions, is that right? Or is it apparent? $\endgroup$ – alex.jordan Dec 22 '17 at 17:30
  • $\begingroup$ @MaxW: I give that solution in the text $\endgroup$ – Ross Millikan Dec 22 '17 at 18:45
  • $\begingroup$ @alex.jordan: it seems apparent to me that $x^3+y^3=30$ has no solution, just because I know the small cubes. It does need to be checked, but since 2^3 \lt 15 you just have to check $y=3$ and then note that $x^3=3$ has no solution. $\endgroup$ – Ross Millikan Dec 22 '17 at 18:47
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Since $57$ is odd, we need $x+y+z$ odd.
Since $3\mid 57$ and $n\equiv n^3\bmod 3$, we need $3\mid x+y+z$.
Given that $-5\le x,y,z\le5$, it's relatively quick to eliminate $x+y+z=\pm 9$ and identify that we need $x+y+z=\pm 3$.

Then taking $x\ge y\ge z$ we must have $x\ge 3$ initially and then after considering a couple of cases we get $x\ge 4$, and we can also quickly eliminate $x=5$ leaving only $x=4$ to explore, which quickly leads to $y=1, z=-2$.

If all permutations are required then there are $3!=6$ arrangements of these values, of course.

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  • $\begingroup$ There are $6$, not $8$ arrangements, but a good approach to reducing the cases. $\endgroup$ – Ross Millikan Dec 22 '17 at 18:48
  • $\begingroup$ @RossMillikan thanks for the check $\endgroup$ – Joffan Dec 22 '17 at 23:03
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Since $57$ is odd, either $x$, $y$ and $z$ are all odd, or just one is odd. It's easy to see that no sum or subtraction of $1$, $27$ and $125$ equals $57$ (the closest you can get is $1+27+27=55$), so we may assume that $x=2u$ and $y=2v$ are even and $z$ is odd.

Looking mod $8$, we have

$$1\equiv57=8u^3+8v^3+z^3\equiv z$$

so we must have $z=1$. This leaves $u^3+v^3=7$ with $|u|,|v|\le2$. We may assume $u$ is even and $v$ is odd. It's easy to see we must have $u=2$ and $v=-1$. (If you like, work mod $8$ again: $-1\equiv7\equiv v^3\equiv v$ mod $8$.) Thus $(x,y,z)=(4,-2,1)$ and its permutations comprise all solutions with absolute values less than or equal to $5$.

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