<s>[INST] Find $a, b, c$ such that $a^3+b^3+c^3-3abc=2017$. 
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?
 [/INST] $$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2} \bigg( (a-b)^2 + (b-c)^2 + (c-a)^2 \bigg)=1$$
$$ \implies (a-b)^2 + (b-c)^2 + (c-a)^2 =2$$
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 $c=673$.
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