You have taken this, right?

This is not a solution you should write on an olympiad.

So when we first see this problem, our first instinct is to do some $\pmod p$ analysis for both sides. $2012$ being a constant gives us something to play around with. For instance, because $\gcd(x,xyz+2) = 1$ , $x^3|2012$ . So if $x$ is even we write $x = 2k$ and divide off by the $2$ to get that $k| 503 \Rightarrow k = 1$ . Otherwise $x$ is odd and $x=1$ .

We split into cases.

Case 1 : $x=1$

so $y^3 + z^3 = 2012 (yz+2)$ . Again we try to find a nice mod (that clearly should divide the more 'fixed' RHS) to work with, influenced by our above analysis, we chose the (albeit a bit daunting) $503$ . So we get the following $y^3 \equiv -z^3 \pmod {503} \Rightarrow y^2 \equiv z^2 \pmod {503} \Rightarrow y+z \equiv 0 \pmod {503}$ . So for simplicity sake let $y+z = 503 \times c$ . The good thing is that we can actually substitute this back into our original equation. I hope you can see where I am getting at. By this substitution, we can bring down the degree of the equation (as we will see we can consider it in terms of $y$ ) and hopefully be able to whack it with more known methods and/or infinite descent or sth of the sort.

So $(3c+4)y^2-(3c+4)503yc+503^2c^3-8 = 0$ . Alright it is quadratic. Anything fancy to do? Uhm. okay we need a way to isolate $t$ and get rid of the annoying $y$ that shouldn't even be around in the first place. asdfghjkl this is annoying. Let us see, we want the coefficients so we should just try whacking the discriminant and factorising the annoying stuff. So...

$503^2 \times -3 c^3+4 \times 503^2 c^2+32\geq 0$

Alright this looks too negative. So we should take a small value of $c$ . Let us establish this rigorously by factorising.

$(c-2)(-3 \times 503^2 c^2-503^2 \times 2 c-4 \times 503^2)+32-8 \times 503^2\geq 0$

So we need $c=1$ . You chuck it in and you realise that you need $108437$ to be a square. But it isn't. So no solutions for this case.

Case 2 : $x= 2$

so $y^3 + z^3 = 503 (yz+2)$ . Once again, as above you should be able to get $y+z = 503c'$ . And we shall apply the same strategy as above. Again subbing this in and doing the calculations (i.e. taking the discriminant), one needs (I might have skipped a bit of the factorisation):

$(c't+1)y^2-503(3c'+1)c'y+ c'^3 503^2-1 = 0 \Leftrightarrow (c'-2)(-503^2c' - 503^2)-2 \times 503^2 + 4 \geq 0$

Once again we need $c' = 1$ . But you get $(x,y,z) = (2, 251, 252)$ as a solution.

So we have exhausted all cases, and your answer is $2+251+252 = \fbox{505}$

P.S. This is longer than an average solution, but should contain enough motivation for this N2.