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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

Tito Piezas III
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user354545
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4 Answers4

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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 $5$.

2'5 9'2
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Ross Millikan
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    you forgot negative integers... -2, 1, 4 work. – MaxW Dec 22 '17 at 16:58
  • 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? – 2'5 9'2 Dec 22 '17 at 17:30
  • @MaxW: I give that solution in the text – Ross Millikan Dec 22 '17 at 18:45
  • @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. – Ross Millikan Dec 22 '17 at 18:47
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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 $(x,y,z)=(-2,1,4)$.

2'5 9'2
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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.

Joffan
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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 $5$.

Barry Cipra
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