Integer solutions to $x^3+y^3+z^3 = x+y+z = 8$

Question

Find all integers $x,y,z$ that satisfy $$x^3+y^3+z^3 = x+y+z = 8$$

Let $a = y+z, b = x+z, c = x+y$. Then $8 = x^3+y^3+z^3 = (x+y+z)^3-3abc$ and therefore $abc = 168$ and $a+b+c = 16$. Then do I just use the prime factorization of $168$?

Answer 1

From $\sqrt[3]{abc}\approx 5.5 > 5\tfrac13=\frac{a+b+c}{3}$, we know that $a,b,c$ cannot all be positive (for in that case the AM-GM inequality states a "$\le $").

As $abc>0$, exactly two of $a,b,c$ must be negative.

Also, either all three numbers are even or exactly two are odd. In the first case, $\frac a2\frac b2\frac c2=21$, so all three integers $\frac a2,\frac b2,\frac c2$ are odd; but then $8=\frac a2+\frac b2+\frac c2$ would be odd as well. We conclude that exactly two of $a,b,c$ are odd, say $a,b$ are odd, $c$ is even. Wlog., $|a|\le |b|$. Then $c$ is an odd multiple of $8$, hence $a+b\equiv 8\pmod{16}$. The only ways to combine numbers from $\pm1,\pm3,\pm7,\pm21$ meeting this are $(a,b)=(\pm1,\pm7)$, or $(a,b)=(\pm3,\pm21)$; but the latter is not compatible with $ab\mid 168$. As two of $a,b,c$ must be negative, we conclude $$a=-1,\quad b=-7,\quad c=16-a-b=24$$ and from this $$ x=\tfrac{b+c-a}2=9,\quad y=15,\quad z=-16.$$

Answer 2

Hint:

Taking from where you left off: $ab \mid 168 \implies ab = \pm 1, \pm 2, \pm 4, \pm 6, \pm 7, \pm 8, \pm 12, \pm 14, \pm 21, \pm 24, \pm 28, \pm 42, \pm 56, \pm 84, \pm 168$. Even though it looks cumbersome, it is easy to solve. For example, $ab = 6 \implies c = \dfrac{168}{6} = 28 \implies a+b = 16-c = 16 - 28 = -12$. This case yields no solution. But other cases yield some solutions as you go through them you might find the work is not that tedious and may be fun indeed.

Answer 3

We may even look out for integer solutions of $x^3+y^3+z^3=8$ alone. There is a huge literature on the sum of three cubes, and several data bases available. The equation has infinitely many solutions, for example $$ \displaystyle (18t^3 + 2)^3 + (18t^4)^3 + (-18t^4 - 6t)^3 = 8 $$ for all integers $t$. Among other solutions we find $$ 15^3+(-16)^3+9^3=8, $$ which qualifies for $x+y+z=8$. Of course, we should solve the question as you did, but at least we already know a solution.