Find $a, b, c$ such that $a^3+b^3+c^3-3abc=2017$.

Question

Find all natural numbers $a, b, c$ such that $a\leq b\leq c$ and $a^3+b^3+c^3-3abc=2017$.

My Attempt

$$a^3+b^3+c^3-3abc=2017$$ $$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=2017*1$$ Now, $a+b+c$ can't be equal to $1$ as $a, b, c$ are natural numbers.
So, $$a+b+c=2017$$ $$a^2+b^2+c^2-ab-bc-ca=1$$ How should I proceed after this?

Answer 1

$$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2} \bigg( (a-b)^2 + (b-c)^2 + (c-a)^2 \bigg)=1$$ $$ \implies (a-b)^2 + (b-c)^2 + (c-a)^2 =2$$

So, two between $a,b,c$ are equal and the other has the difference of $1$ from the others.

WLOG, assume $b=a$ and $c=a \pm 1$ (Ignoring $a\le b\le c$ here.).

$$a+b+c=3a\pm1 = 2017 = 3\times672 \,+1$$

Thus $c=a+1$, $a=b=672$ and $c=673$.

Answer 2

These are surfaces over {a,b} expected reals. The calculations are straightforward.

The problems are the second and third solutions are complex. The graph shows only the real parts.

Still calculations are straightforward.

Answer 3

It was an easy question, just one important result and a bit of number theory was needed