Solve $a^3 -b^3 -c^3=3abc, a^2=2(b+c)$ in natural numbers.

Question

Solve $a^3 -b^3 -c^3=3abc, a^2=2(b+c)$ in natural numbers.

Substituting $a=\sqrt{2(b+c)}$ in the cubic equation, we get:
$2\sqrt{2}(b+c)^{\frac{3}{2}} - b^3 -c^3 = 3\sqrt{2(b+c)}bc$
$2\sqrt{2}(b+c)\sqrt{b+c} - b^3 -c^3 = 3\sqrt{2(b+c)}bc$

Not able to proceed further, although I think that binomial expansion of $\sqrt{b+c}$ can yield some more steps. But, am confused about applying that too.

Answer 1

\begin{align}x^3+y^3+z^3-3xyz & = x^3+y^3+3x^2y+3xy^2+z^3-3xyz-3x^2y-3xy^2 \\ & =(x+y)^3+z^3-3xy(x+y+z)\\ &= (x+y+z)((x+y)^2+z^2-(x+y)z)-3xy(x+y+z) \\ & =(x+y+z)(x^2+2xy+y^2+z^2-xy-xz-3xy) \\ & =(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\end{align}

Therefore, for distinct $x,y $ and $z$; $$ x^3+y^3+x^3=3xyz \implies x+y+z=0$$ Notice that your equation can rewritten as $$(a)^3+(-b)^3+(-c)^3=3 (a) (-b)(-c)$$

$$\implies a-b-c=0$$

Substitute $a=b+c$ in your second equation,

$$ a^2=2(b+c) \implies (b+c)^2=2(b+c) \implies b+c=2$$

So finally , we've $a=2$ and $b+c=2$.

So the only natural solution to these equations is $\color{blue}{(a,b,c)=(2,1,1)}$.

Answer 2

Hint:
$$0=a^3-b^3-c^3-3abc = \frac12(a-b-c)[(a+b)^2+(b-c)^2+(c+a)^2]$$ So we have two cases: Case 1:
$a=b+c \implies a^2=2a \implies a \in \{0, 2\}$.
You should be able to fnd out $b+c$ bounded in this case...

OR Case 2:
$a+b=b-c=c+a=0$, which again reduces cases considerably among natural numbers.