First we see $503\not| x$ and then $x\in\{1,2\}$. Now $503|(y+z)(y^2-yz+z^2)$. Clearly $503\not|yz$ If $503|y^2-yz+z^2$ then $503|(2y-z)^2+3z^2$ so $-3$ is square with respect to module $503$ which is not true. Then $y+z= 503t$. So we get: \[x^3t(y^2-y(503t-y)+(503t-y)^2)= 4xy(503t-y) +8\] If $x=2$ we have quadratic equation ith respect to $y$: \[y^2(3t+1)-503ty(3t+1)+503^2t^3-1=0\] and it's discriminant is \[ D= (3t+1)(4+503^2t^2-503^2t^3)\] and ince $D\geq 0$ we have $ t^3\leq {4\over 503^2} +t^2$ which is possible only for $t=1$. Then $D=16$ and $y_{1,2}={2012\pm 4\over 8}$. If $y= 251$ then $z=252$ and if $y=252$ then $z=251$. If $x=1$ we procedue similary (I suppose)...

Number1 wrote: If $x=1$ we procedue similary (I suppose)... Yes, we can, but it's little bit more bothersome;

Similar to the above post, we have $503 \nmid x \implies 503|y^3+z^3$. Since $( \frac{-3}{503} )=-1$, we must have $503|y+z$. We must have $x^3|4(xyz+2) \implies x|8 \implies x \in { 1,2 }$. Now, the way I did is much less tedious: Case I: $x=1$. If $y$ and $z$ have different parity, $LHS$ is odd but $4|RHS$, contradiction. So, $2|y-z$. If $(y,z)=1$, then they are both odd implying $y^2-yz+z^2$ and $yz+2$ are odd. Therefore, $2012|y+z \implies 2012 \le y+z \implies y^2-yz+z^2 \le yz+2$ giving $(y-z)^2 \le 2$. So, $y-z=1$ giving. If possible, let $p$ be a prime dividing $y$ and $z$. Then $p|p^3|LHS=503.4.(yz+2)$. Obviously, $p \neq 503$. Then we have $p=2$. If $y=pa$ and $z=pb$, the equation transforms into $a^3+b^3=503(2ab+1)$. So, $a$ and $b$ have different parity and hence, $a+b$ is odd. If $a+b \ge 503.3$, we get an obvious contradiction. Otherwise, $a+b=503 \implies a^2-3ab+b^2=1$. Its integer solutions may be easily found and case checked to give solutions. Case II: $x=2$. $4(z^3+y^3)=2012(yz+1) \implies z^3+y^3=503(yz+1)$. Since $503|y+z$, we have $y^2-yz+z^2 \le yz+1 \implies y-z \le 1$. Therefore, $y=z$ or $y=z+1$ which give single variable equations.

As in the previous posts, I will use the fact that $x=1, 2$ and that $503\mid y+z$. Now, if $x=2$ we have \[\frac{y+z}{503}=\frac{yz+1}{y^{2}-yz+z^{2}}\in \mathbb{Z}^{+}\] so $y^{2}-yz+z^{2}\leq yz+1$ and therefore $\left(z-y\right)^{2}\leq 1$. If $y=z$ then $y^{2}\mid y^{2}+1$ so $y=z=1$ which is a contradiction since $503\nmid 2$. If $y+1=z$ then we have \[\frac{2y+1}{503}=\frac{y^{2}+y+1}{y^{2}+y+1}=1\] so $2y+1=503$ and $y=251$. Thus, the triplet $\boxed{\left(2, 251, 252\right)}$ works for the problem's conditions and we are done with the case $x=2$ because $z-y<2$. Now, for the case $x=1$ we proceed in a similar way. We will have $y+z$ must be even, so \[\frac{y+z}{1006}=\frac{2\left(yz+2\right)}{y^{2}-yz+z^{2}}\in \mathbb{Z}^{+}.\] We check the cases $y=z$ and $y+1=z$. For the first case, we get $y^{2}\mid 2y^{2}+4$, so $y\mid 2$ and therefore $y+z\leq 4<503$ which is a contradiction to the fact that $503\mid y+z$. For the second case, we get $y^{2}+y+1\mid 2\left(y^{2}+y+2\right)$, so $y^{2}+y+1\mid 2$ which clearly is not possible. Hence, $\left(z-y\right)^{2}>2$ and with this we get that \[2\left(y^{2}-yz+z^{2}\right)>2\left(yz+2\right)\geq y^{2}-yz+z^{2}\] and with this, we see that it must hold that $y^2-yz+z^{2}=2yz+4$ and $y+z=1006$. Let $y=503-d$ and $z=503+d$, then we must have that \[\left(z-y\right)^{2}=4d^{2}=yz+4=\left(503-d\right)\left(503+d\right)+4=503^{2}+4-d^{2}\] or equivalently, $5d^{2}=503^{2}+4$ wich is not possible since $5\nmid 503^{2}+4$. Hence, there are no solutions for the case $x=1$ and thus the only solution is \[\boxed{\left(2, 251, 252\right)}\]

If $503|x$ then $503^3$ divides the LHS while $503$ is the largest power of $503$ that divides the RHS. Therefore $503\not{|} \ x$. Thus, since $x|2012,$ we must have $x=2^a$ where ${a\in\{0,1,2}$. If $x=2^2$ then $2^6$ divides the LHS while the largest power of $2$ that divides the RHS is $2^3$ therefore $x=\{1,2\}$. Additionally, since $503$ is prime, we have \begin{align*}y^{502}&\equiv z^{502}\pmod{503}\\ y^3&\equiv -z^3\pmod{503}\\ y^{3\cdot 167}&\equiv -z^{3\cdot 167}\pmod{503}\\ \implies y&\equiv -z\pmod{503} \end{align*} Therefore $503|y+z$ and thus $y+z=503k$. CASE 1: $x=1$ Then the equation becomes \begin{align*} y^3+z^3&=2012(yz+2)\\ 503k(y^2-yz+z^2)&=2012(yz+2)\\ k(y^2-yz+z^2)&=4yz+8\\ k(y-z)^2+yz(k-4)&=8\\ \end{align*} Since $yz(k-4)\leq 8$, we must have $k\leq 4$ since $yz+1\geq y+z=503k\geq 503\implies yz\geq 502$. Additionally, since $y^3+z^3\&=2012(yz+2)$, we must have $y^3+z^3$ to be even and thus $y$ and $z$ have the same parity making $y+z$ be even as well. But $y+z=503k$ so $k$ is even. Therefore, $k\in\{2,4\}$. If $k=4$ then $(y-z)^2=2$ which is impossible. If $k=2$ then $2(y-z)^2-2yz=8\implies (y+z)^2-5yz=4\implies 2^2\cdot 503^2-4=5yz$ but the LHS is not divisible by $5$ therefore there is no solution when $x=1$. CASE 2: $x=2$ The equation then becomes \begin{align*}y^3+z^3&=503(yz+1)\\ k(y^2-yz+z^2)&=yz+1\end{align*} However, if $|y-z|> 1$, then $k(y^2-yz+z^2)\geq y^2-yz+z^2> yz+1$ therefore $y=z$ or $|y-z|=1$. The former produces $2y^2=503y^2+503$ which makes $503|y$ and so $503^2||2y^2$ and $503||503y^2+503=2y^2$, a contradiction. The latter makes $k=1$ and produces the solution \[(x,y,z)=\boxed{(1,251,252)}.\]

JSGandora wrote: \[(x,y,z)=\boxed{(1,251,252)}.\] It must be $(2,251,252)$

Lemma 1: $ 503 \nmid x(y^2 - yz + z^2) $ Proof: Note that $ 503 \vert x \Longrightarrow 503^3 \vert 2012(xyz + 2) \Longrightarrow 503 \vert (4xyz + 8) $ but clearly $ 4xyz + 8 \equiv 8 \pmod{503} $, contradiction! Moreover, $ 503 \vert y^2 - yz + z^2 \Longrightarrow 503 \vert (2y - z)^2 + 3z^2 \Longrightarrow \left(\frac{-3}{503}\right) = 1 $. Since $ 503 \equiv -1 \pmod{4} \Longrightarrow \left(\frac{-1}{503}\right) = -1 $, this means that $ \left(\frac{3}{503}\right) = -1 $. But by Quadratic Reciprocity we have $ \left(\frac{3}{503}\right)\left(\frac{503}{3}\right) = -1 $ and since $ \left(\frac{503}{3}\right) = \left(\frac{2}{3}\right) = -1 $, we know that $ \left(\frac{3}{503}\right) = 1 $, contradiction! Lemma 2: $ x \vert 2 $ Proof: Note that $ x^3(y^3 + z^3) = 2012(xyz + 2) \Longrightarrow x \vert 4012 $ but since $ (503, x) = 1 $ by Lemma 1, this actually implies that $ x \vert 8 $. But if $ 4 \vert x $, then $ v_2(2012(xyz + 2)) = 3 < v_2(x^3(y^3 + z^3)) $, contradiction! Note that Lemma 1 implies that $ 503 \vert (y + z) $. Now, we proceed with casework. Case 1: $ x = 1 $ Our equation becomes $ (y + z)(y^2 - yz + z^2) = 2012(yz + 2) $. Since the RHS is even, we must have $ y \equiv z \pmod{2} $. Moreover, since by AM-GM we have $ (y + z)(y^2 - yz + z^2) \ge (y + z)yz $ it is clear that $ y + z < 2515 $. This means that either $ y + z = 1006 $ or $ y + z = 2012 $. If $ y + z = 2012 $ then the equation can be written as $ (y - z)^2 = 2 $ which clearly has no integer solution. If $ y + z = 1006 $ then the equation becomes $ y^2 - 3yz + z^2 = 4 $. But since in this case $ y + z = 1006 \Longrightarrow y^2 + 2yz + z^2 = 1006^2 $ this means that $ 5yz = 1006^2 - 4 $ which also clearly has no integer solution. Case 2: $ x = 2 $ Our equation becomes $ (y + z)(y^2 - yz + z^2) = 503(yz + 1) $. Since by AM-GM $ (y + z)(y^2 - yz + z^2) \ge (y + z)yz $ it's clear that we must have $ y + z = 503 $. Then the equation becomes $ (y - z)^2 = 1 $ and since we are given that $ y \le z $ we obtain the unique solution $ \boxed{(x, y, z) = (2, 251, 252)} $.

I dont think mine is a better or newer solution but still here it goes: Firstly we notice that $x=1$ or $x=2$(as in previous solutions only trivial arguments) Case 1 If x=1 Clearly, we let $y+z=503k$ and $y^2-yz+z^2=\frac{4(yz+2)}{k}$ since we note that if $503 \mid y^2-yz+z^2$ then by quadratic reciprocity $503 \equiv 1\bmod 3$ false indeed. Now, $yz=\frac{503^2k^3-8}{3k+4}$ and by using $(y+z)^2 \ge 4yz$ we get that $k \le 4$. Moreover, $3k+4 \mid 503^2k^3-8$ giving a contradiction for $k=1,2,3,4$ do not satisfy this. Case 2 If x=2 Again, we get that $y+z=503t$ and $y^2-yz+z^2=\frac{yz+1}{t} \ge yz$. Giving $t=1$ and $y-z=+1/-1$. Now, this yields $(2,251,252);(2,252,251)$ as the only solutions.$\square$.

Hmm, I actually feel like showing $x=1, 2$ is a large part of the problem. Anyways, here's another solution. So we obviously have $503|x, 503|(y+z)$, or $503|(y^2-yz+z^2)$. Now casework Case 1: $503|x$ We have $x=503k$, so then $503^2 k^3 (y^3+z^3) = 4(503kyz+2)$, contradiction by $\pmod{503}$. Case 2: $503|(y+z)$ Write $y+z=503k, k \ge 1$. Rewrite the equation as $kx^3(y^2-yz+z^2) = 4(xyz+2) \ge kx^3yz \implies 8 \ge yz(kx^3-4x) \ge x^3 (x^2-4) \implies x = 1, 2$. If $x=1$, then the equation becomes $k(y^2-yz+z^2)=4(yz+2)$. If $yz \le 8, y+z \equiv 0 \pmod{503}$ cannot be satisfied, so by AM-GM we get $k \le 4$. Now, rewrite the equation as $k(y+z)^2 = yz(3k+4)+8$. Casework on $k=1, 2, 3, 4$ all yield contradictions for this case. If $x=2$, then the equation becomes $k(y^2-yz+z^2)=yz+1 \leftrightarrow k(y+z)^2 = yz(3k+1)+1$. By AM-GM, we have $kyz \le yz+1$, so $k=1$. Then $y-z=1 \implies y=251, z=252$. Case 3: $503|(y^2-yz+z^2)$ So $y^2 - yz + z^2 \equiv 0 \pmod{503}$. Since $\gcd (4, 503)=1$ we can rewrite the congruence in this case as \[ (2y-z)^2 + 3z^2 \equiv 0 \pmod{503} \implies \left(\frac{-3}{503} \right)= 1 \]$503 \equiv 3 \pmod{4} \implies \left(\frac{-1}{503}\right) = -1$, so $\left(\frac{3}{503}\right) = -1$. But by QR Law, \[ \left(\frac{3}{503}\right) \left(\frac{503}{3}\right) = -1 \implies \left(\frac{3}{503}\right) = 1 \]because $\left(\frac{2}{3}\right) = -1$. Contradiction! So our only solution is $\boxed{2, 251, 252}$.

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I am pretty surprised that no one seems to have used the lemma that if $p$ is an odd prime dividing $x^2 -xy +y^2$ where x and y are integers with x y integers not being divisible by p , then p cannot be 2mod3.... with this it breaks into casework....and you get the answer.....

Can you post the solution.

kk108 wrote: I am pretty surprised that no one seems to have used the lemma that if $p$ is an odd prime dividing $x^2 -xy +y^2$(you probably mean -instead of +) where x and y are integers with x y integers not being divisible by p , ten p cannot be 2mod3.... with this it breaks into casework....and you get the answer.... In fact every single solution i see, used this.

Thank you for correcting my reading skills! I am not familiar with this Legendre symbol by the way... Anyhow your post strengthened my conviction in my solution being right! Thanks!

Can anybody provide me good article on quadratic reciprocity and legendre symbol.

There was a pretty good article --- if you type something like that in the search in the HSO forum to get it.

lyukhson wrote: Find all triples $(x,y,z)$ of positive integers such that $x \leq y \leq z$ and \[x^3(y^3+z^3)=2012(xyz+2).\] Here is my solution. First we present the following: Lemma. Let $x,y$ be integers, $p>2$ a prime number, and $k$ a positive integer relatively prime with $p-1.$ If $p$ devide $x^k+y^k,$ then $p$ devide $x+y.$ Proof. Since $k$ and $p-1$ are relatively prime, then there are two integers $u,v$ such that $ku-v(p-1)=1.$ Assume that $p$ devide $x^k+y^k.$ If $p$ devide one of $x^k,y^k$ say, $x^k$ then $p$ devide $y^k$ and thus, $p$ devide both of $x,y,$ which give us $p\mid x+y.$ If $p\nmid x^k,y^k,$ then $\gcd(p,x)=\gcd(p,y)=1,$ and from Fermat' little theorem, we have $x^{p-1}\equiv y^{p-1}\equiv 1\pmod p.$ We deduce that $$x\equiv x^{ku-v(p-1)}\equiv (x^k)^u (x^{p-1})^{-v}\equiv -(y^k)^u(y^{p-1})^{-v}\equiv -y^{ku-v(p-1)}\equiv -y \pmod p,$$as wanted. Back to the problem. Let $(x,y,z)$ be a triple from positive integers satisfy the conditions of the problem. We see that $2012=2^2\times 503,$ so we consider two cases: If $x$ is even then there is a positive integer $k$ such that $x = 2k,$ and the equation become $k^3(y^3+z^3) = 503(kyz+1).$ Because $\gcd(k^3,kyz+1) = gcd(k, kyz+1) = 1,$ we deduce that $k^3$ must devide $503$ and thus, $k = 1$ and $x = 2.$ The equation become $y^3 +z^3 =503(yz +1).$ Beacuse, $(503-1)-3\times 167=1,$ and $503$ devide $y^3+z^3,$ we deudce from the lemma that $503$ devide $y+z.$ Therefore, $y^2-yz+z^2$ devide $yz+1.$ But, if $z>y+1$ then \[y^2-yz+z^2=(z-y)^2+yz>yz+1,\]which is rejected. Hence, $z=y,y+1.$ If $z=y,$ then $y^2$ devide $y^2+1$ and this give us $z=y=1,$ which is rejected $(z\ge y\ge x=2).$ If $z=y+1,$ then the equation can be written $503=y+z=2z+1.$ Therefore, in this case $(x,y,z)=(2,250,251).$ If $x$ is odd then $x^3$ and $xyz+2$ are relatively prime, and thus $x^3$ must devide $2012,$ which means that $x = 1.$ The equation become $y^3+z^3= 2012(yz+2).$ Again, we deduce from the lemma that $503$ devide $x+y.$ The equation become $k(y^2-yz+z^2)=4(yz+2),$ where $k$ is a positive integer. We consider two subcases: If both $y$ and $z$ are even, then there are two positive integer $y_1,z_1$ such that $y=2y_1$ and $z=2z_1.$ The equation become $k(y_1^2-y_1z_1+z_1^2)=y_1z_1+2.$ Since, $$y_1^2-y_1z_1+z_1^2=(z_1-y_1)^2+y_1z_1\ge y_1z_1+4>y_1z_1+2$$when $z_1\ge y_1+2,$ we deduce that $z_1=y_1,y_1+1.$ If $y_1=z_1,$ then $z_1^2$ devide $z_1^2+2,$ hence, $z^2$ devide $2,$ and we get that $z_1=1,$ and $y=z=2,$ which is rejected $(1(2^3+2^3)<2012(4+1)).$ If $z_1=y_1+1$ then, $y_1z_1+1>1$ devide $y_1z_1+2$ but these two numbers are relatively prime, hence this case is rejected. If one of $y,z$ is odd and the other is even, or both of them are odd, then $y^2-yz+z^2$ is odd and thus, $yz+2$ is a multiple of $y^2-yz+z^2.$ Like in the first subcase, this is true only when $y=z=1,$ which is rejected $(1(1+1)<2012(1+2)).$. We conclude that this equation has only one solution, which is $(x,y,z)=(2,250,251).$

lyukhson wrote: Find all triples $(x,y,z)$ of positive integers such that $x \leq y \leq z$ and \[x^3(y^3+z^3)=2012(xyz+2).\] Firstly $2012 = 2^2 \cdot 503$ and 503 is prime. So $503 \mid x^2$ or $503 \mid y^3+z^3$ If $503 \mid x^2$, then $(503k)^2(y^3+z^3) = 2012(xyz+2)$ But $(503k)^2 > 2012$ and $y^3+z^3\ge x(y^2+z^2) \ge 2xyz > xyz+2$ so no solutions. Crucial claim: $503\mid y^3+z^3$, then $y^3 \equiv z^3 \pmod{503}$, so \begin{align*}(-z)^{502}&\equiv1\equiv y^{502} \equiv y(-z)^{501}\\ \implies& y \equiv -z \pmod{503}\\ \implies& 503 \mid y+z \end{align*}Let $503k=y+z$. Then $2012(xyz+2) = x^3(y+z)(y^2-yz+z^2) = 503kx^3(y+z)$ so $4(xyz+2) = kx^3(y+z)$ rearranging to $$kx^3(y-z)^2+yz(kx^3-4x)=8$$If $x \ge 3$, $kx^3-2x \ge 15$ so no solutions. If $x = 2$, $k(y-z)^2+yz(k-1)=1$ $k \ge 2$: $yz(k-1) \ge 1$ so no solutions $k = 1$: $(y-z)^2=1 \implies (y,z)=(251,252)$ If $x=1$: $k(y-z)^2+yz(k-4)=8$. We do a long case bash: $k \ge 5$: $yz(k-4) > 8$ so no solutions $k = 4$: $4(y-z)^2=8$ but solutions not integer $k = 3$: $3(y-z)^2=8+yz$. Since $y+z = 503 \cdot 3$, $yz$ is even, RHS is even, but $y-z$ odd so LHS is odd, no solutions $k = 2$: $2(y-z)^2=2yz+8 \implies (y-z)^2 = yz + 4$. Then $1006^2=(y+z)^2 = 5yz + 4$, but $1006^2 -4$ not divisible by 5, so no solutions $k = 1$: similar $k=3$ Therefore $(x,y,z)=(2,251,252)$.

A pretty clean solution, imo. If $x=y=1,$ then $1+z^3=2012(z+2) \implies z(z^2-2012)=4023$ has no solutions. If $y>1,$ then $x^3(y^3+z^3)=2012(xyz+2) \leq 2012(z^3+2) \leq 2012(z^3+y^3) \implies x^3 \leq 2012 \implies x \leq 12.$ Now, taking the equation mod $x,$ we get $4024 \equiv 0 \pmod{x},$ so $x=2,4,8.$ If $x=4,$ we get $16(y^3+z^3)=503(4yz+2),$ but LHS is 0 mod 16, RHS is 2 mod 4. If $x=8,$ we get $8^3(y^3+z^3)=2012(8yz+2),$ which again is not possible by looking at $v_2$ of each side. Thus, $x=2$ gives us $\boxed{y^3+z^3=503(yz+1)} \implies (y+z)(y^2-yz+z^2)=503(yz+1).$ If $y=z,$ we get $ 2y^3-503y^2=503,$ so $y^2(2y-503)=503,$ gut $503$ is prime and we get no solutions. For $z>y>1,$ we have $(y-z)^2 \geq 1 \implies y^2-yz+z^2 \geq yz+1,$ so that means $y+z \leq 503.$ Claim: If $a^3 \equiv b^3 \pmod{503},$ then $a \equiv b \pmod{503}.$ Proof: Consider $ab^{-1}.$ This has either order one or order three. However, the order must divide $\varphi(503)=502,$ so it has order one, meaning $a \equiv b \pmod{503} \quad \square$ Thus, $y^3+z^3 \equiv 0 \pmod{503} \implies y^3 \equiv -z^3 \equiv (-z)^3 \pmod{503},$ which means $y \equiv -z \pmod{503}.$ Thus, $y+z=503,$ and our factored equation cancels to $y^2-yz+z^2=yz+1 \implies (y-z)^2=1,$ so $y=251,z=252.$ Our only solution is $\boxed{(2,251,252)} \quad \blacksquare$

Obey me, bounding! Obey me, orders! We claim that the only solution is $(2,251,252)$. We can verify that it works. Note that $x\mid 2012xyz+4024,$ implying $x\mid 4024.$ We claim that $503\nmid x.$ Note that $503^3\mid x^3(y^3+z^3)$ but $503^3\nmid 2012(xyz+2),$ since $xyz+2\equiv 2\pmod{503}.$ Now we examine $\nu_2{x}.$ Note that $4\nmid x$ otherwise $4^3\mid x^3(y^3+z^3)$ but the largest power of $2$ that divides $2012(xyz+2)$ is $2^3.$ So we're restricted to $x=1$ and $x=2.$ Note $x=1$ is equivalent to $y^3+z^3=2012(yz+2)$ and $x=2$ is equivalent to $y^3+z^3=503(yz+1).$ We claim that $y^3+z^3=2012(yz+2)$ has no solutions. Note that $y^3\equiv(-z)^3\pmod{503},$ and since $3\nmid \phi(503),$ $y\equiv -z\pmod{503}.$ Similarly, $y\equiv -z\pmod{4}.$ Thus $2012\mid y+z.$ Now we bound. Say $y+z=k.$ Then $y^3+z^3\geq \frac{k^3}{4}$ and $2012(yz+2)\leq 2012(\frac{k^2}{4}+2).$ Clearly for $k>2012$ the equation has no solutions, since for $k\geq 4024,$ $y^3+z^3\geq \frac{k^3}{4}\geq 2012(\frac{k^2}{2})\geq 2012(\frac{k^2}{4}+2).$ So the only possible solution is $y+z=2012.$ Now define $f(z)=z^3+(2012-z)^3$ and $g(z)=2012(z(2012-z)+2).$ Note that on the interval $1006\leq z\leq 2011,$ $f(z)$ is increasing and $g(z)$ is decreasing. Notice that $z=1006$ obviously doesn't work and that $f(1007)>g(1007),$ so there are no solutions in this case. The next case proceeds identically. Note that $503\mid y+z$ by the same reasoning as above, and say $y+z=k.$ Clearly for $k>503$ the solution has no solutions, since for $k\geq 1006,$ $y^3+z^3\geq \frac{k^3}{4}\geq 503(\frac{k^2}{2})\geq 503(\frac{k^2}{4}+1).$ Now let $f(z)=z+(503-z)^3$ and $g(z)=503(z(503-z)+1).$ Clearly $f(z)$ is increasing and $g(z)$ is decreasing along the interval $252\leq z\leq 501,$ and $f(252)=g(252),$ so we are done.

ISL 12N2. Find all $(x, y, z)\in\mathbb{N}^3$ such that $x\le y\le z, x^3(y^3+z^3)=2012(xyz+2)$. Solution 1. First off notice that $x\mid 4024=2^3\cdot 503$. However if $503\mid x$ then $1=v_{503}(2012(xyz+2))=v_{502}(x^3(y^3+z^3))\ge 3$, which is a contradiction. Thus $x\in\{2^k: 0\le k\le 3\}$. Also, $3\ge v_2(2012(xyz+2))=x_2(x^3(y^3+z^3))\ge 3v_2(x)\implies x\in\{1, 2\}$. Now it is obvious that $503\nmid y, 503\nmid z$ and $503\mid y^3+z^3$, which implies that $y^{504}=z^{504}$. We have from FLT that $y^{502}\equiv 1\equiv z^{502}\pmod{503}\implies y^2\equiv z^2\pmod{503}$. Now the prime $503\mid (y+z)(y-z)$. If $503\mid y-z$ then $z^3\equiv y^3\equiv -z^3\pmod{503}$, which is a contradiction, hence $503\mid y+z$. We now write $k=\frac{y+z}{503}$ and analyse the situation by dividing into two cases: Case 1. $x=2$ The equation becomes $y^3+z^3=503(yz+1)\iff k(y^2-yz+z^2)=yz+1$. However, by AM-GM or whatever, $y^2-yz+z^2\ge yz\implies k=1$ (note that $y, z$ are large enough). From here we get $(y-z)^2=1$ and so $y-z=-1$. Therefore from $y+z=503$ we get $(x, y, z)=(2, 251, 252)$. Case 2. $x=1$ The equation becomes $y^3+z^3=2012(yz+2)\iff k(y^2-yz+z^2)=4(yz+2)$. Again, by AM-GM or whatever, $y^2-yz+z^2\ge yz\implies k\le 4$ (note that $y, z$ are large enough). Note that $2\mid y^3+z^3\implies 2\mid y+z\implies 2\mid\frac{y+z}{503}=k\implies k\in\{2, 4\}$ since $2\mid y^2-yz+z^2\implies 2\mid y, 2\mid z$. Hence we do casework. Case 2.1. $k=2$ The equation becomes $y^2-yz+z^2=2yz+4\iff (y+z)^2-5yz=4$. However, $5yz=1006^2-4\not\equiv 0\pmod 5$, which is a contradiction. Case 2.2. $k=4$ The equation becomes $y^2-yz+z^2=yz+2$. However this turns out immediately imply an absurd equation $(y-z)^2=2$. On the other hand if $(x, y, z)=(2, 251, 252)$, plugging into the original Diophantine we get both sides equal $254530072$, this shows that this triple is indeed a solution to the Diophantine. In conclusion, the only solution is $(x, y, z)=(2, 251, 252)$. Solution 2. To prove that $503\mid y+z$ from $503\mid y^3+z^3$ (and $503\nmid y, 503\nmid z$) we proceed in another way. Assume on the contrary that indeed $503\mid y^2-yz+z^2$, then $503\mid 4y^2-4yz+4z^2=(2y-z)^2+3z^2\implies -3\equiv ((2y-z)\cdot z^{-1})^2\pmod{503}$. So $-3$ is indeed a quadratic residue modulo $503$. This is absurd, since by the Law of Quadratic Reciprocity, \[\left(\frac{-3}{503}\right)=\left(\frac{-1}{503}\right)\left(\frac{3}{503}\right)=\left(\frac{3}{503}\right)=-\left(\frac{503}{3}\right)(-1)^{\frac{3-1}{2}\cdot\frac{503-1}{2}}=\left(\frac{503}{3}\right)=\left(\frac23\right)=-1.\] We do everything left as in Solution 1.

We claim that the only solutions to the equation are $(x,y,z)=(2,251,252),(2,252,251)$ Factoring gives $x^3(y^3+z^3)=4 \cdot 503 (xyz+2)$ where $503$ is a prime. Now Claim 1:- $x=2$ Proof:- Taking $\pmod x$ gives $x^3(y^3+z^3) \equiv 2012(xyz+2) \equiv 2012 \cdot 2 \equiv 4024 \pmod x$ implying $x|4024$. Now if $503|x$ then $\nu_{503}(x^3+y^3)=1$,which is impossible so $503 \nmid x$ So it suffices just to check $x=1,2,4,8$ and by similar arguments $\nu_2(x)=1$ or $x=2$ By this,$8(y^3+z^3)=2012(2yz+2)=>(y^3+z^3)=503(yz+1)$ Now Corollary 1:- If $p$ is a prime of the form $3K+2$ then the set $[1^3,2^3,........,(p-1)^3]$ is a permutation of $[1,2,3,4,5,6.......,(p-1)]$ modulo $p$ By the Corollary,if $503|y^3+z^3$ it implies that $y+z=503k$ So the equation becomes $(y+z)(y^2-yz+z^2)=503(yz+1)$ By symmetry assume that $y>z$ If $k \ge 2$,then $kyz+1>3yz+1>2(y^2-yz+z^2)=2y^2+2z^2>4z^2$,absurd unless $z=1$(which isn't possible) so $k=1$ and again factoring out gives $y+z=503,y-z=1$ whose solutions are $(x,y,z)=(2,251,252),(2,252,251)$

No MONT solvers here???

What is mont ???????

Amoonguss wrote: What is mont ??????? Modern Olympiad Number theory

Taking $\mod x$ we get that $x\mid 4024$. Notice that if $x\ge 503$ we arrive at a contradiction since $503^3>2012$ and $z^3\ge xyz$ and $y^3>2$. Thus $x=1, 2, 4, 8$. If $x\ge 4$ then $\nu_2(x^3(y^3+z^3))\ge 6$ while $\nu_2(2012(xyz+2))\le 3$. Thus $x=1, 2$. Claim 1. $y+z\equiv 0\pmod{503}$ Proof. We have \begin{align*} y^3\equiv -z^3\pmod{503}\\ y^{502}\equiv z^{502}\pmod{503}\\ y^{501}=(y^3)^{167}\equiv (-z^3)^{167}=-z^{501}\pmod{503} \end{align*}Now multiplying both sides by $y$ in the last congruence we have $y^{502}\equiv -z^{501}\cdot y \pmod{503}$ but because of the second congruence we have $y+z\equiv 0\pmod{503}$. $\blacksquare$ Case 1. $x=1$ We have $(y+z)(y^2-yz+z^2)=2012(yz+2)$. Let $y+z=503k$. Then since $y, z$ have the same parity, $k$ must be even. Then \[ k((y-z)^2+yz)=4yz+8\implies k(y-z)^2+yz(k-4)=8. \]Since $yz\ge y+z-1\implies yz\ge 502$ we must have $k\le 4$. This means that $k=2, 4$. Case 1.1 $k=2$ We then have $y+z=1006$. Thus $2(y-z)^2-2yz=8\implies (y+z)^2-5yz=4$ which is a contradiction since $1006^2-4\equiv 2\pmod{5}$. Case 1.2 $k=4$ We then have $y+z=2012$. Then $y^2-2yz+z^2=2\implies (y-z)^2=2$ which is a contradiction. Case 2. $x=2$ We get $y^3+z^3=503(yz+1)$. Let $y+z=503k$ and we get $k(y^2-yz+z^2)=yz+1$. Then $k(y-z)^2+yz(k-1)=1$. This means that $k\le 1$ or that $k=1$. Then we get $y-z=-1$ and then we find the solution $(2, 251, 252)$. Thus our only solution is the ordered triplet $(2, 251, 252)$.

:sussusamongus: Taking everything mod $x,$ we get $4024\equiv 0\pmod x$ so $x|4024.$ Now, $y^3+z^3=(y+z)(y^2+z^2-yz)\ge 2xyx$ so $x^3(y^3+z^3)\ge 2x^3yz.$ Thus, $2012x(yz)+4024\ge2x^3(yz)$ which implies $(2012x-2x^3)yz\ge 4024.$ Clearly, $2012x-2x^3$ is positive so $1006>x^2,$ implying $x=1,2,4$ or $8.$ Now, if $x=4$ or $8$ then $v_2(x^3(y^3+z^3))\ge 3v_2(x)\ge 6$ and $v_p(xyz+2)=1$ so $v_p(2012(xyz+2))=3.$ Thus, there is a contradiction. If $x=2$ then $y^3+z^3=503yz+503.$ Thus, $503|y^3+z^3.$ We prove that $503|y+z.$ Let $a=\frac{y}{z}.$ $y^3\equiv -z^3 \pmod{503}$ so $a^3\equiv -1\pmod {503}$ so $\text{ord}_{503}(a)=1$ or $3.$ Order must divide $502$ so the order is $1.$ Hence, $503|y+z.$ Now, if $y+z=503k$ then $503k(y^2+z^2-yz)=503(yz+1)$ so $yz+1\ge y^2+z^2-yz$ so $|y-z|\le 1.$ If $y=z$ then $y=z=503m\implies 503m^3=503(m^2+1)$ which gives no solutions. If $y+1=z$ then the only solution is $(2,251,252).$ if $x=1$ then $k(y^2+z^2-yz)=4(yz+2).$ Note that $yz\ge y+z-1\ge 502$ so by simple bounding, $k\le 4.$ None of the values of $k$ give solutions, so we are done.

Note that the prime factorization of $2012$ is $2^2\cdot503$. Claim: $x\in\{1,2\}$ If $503\mid x$, then, where, $x=503a$: $$503^2a^3(y^3+z^3)=2012ayz+8,$$impossible for$\pmod{503}$ reasons. Then $x\mid 8$. Taking $v_2$ of both sides, we have: $$3v_2(x)+v_2(y^3+z^3)=2+v_2(xyz+2).$$If $v_2(x)\ge2$, then the $\text{LHS}$ is at least $3v_2(x)\ge6$ whereas the $\text{RHS}$ is $2+1=3$, contradiction. Hence $v_2(x)=0$ or $v_2(x)=1$, which, combined with $x\mid 8$, means $x\in\{1,2\}$. Now we can conclude that $503\mid y^3+z^3$, so: $$y\equiv y^{1005}\equiv(y^3)^{335}\equiv(-z^3)^{335}\equiv-z^{1005}\equiv-z\pmod{503}$$by FLT, or $503\mid y+z$. Case 1: $x=2$ From $x=2$, the equation rearranges as. $$\frac{y+z}{503}=\frac{yz+1}{y^2-yz+z^2}.$$Then $\frac{yz+1}{y^2-yz+z^2}\ge1$, so $(y-z)^2\le1$. Since $y\le z$, either $y=z$ or $y+1=z$. If $y+1=z$, then $\frac{y+z}{503}=\frac{yz+1}{y^2-yz+z^2}=1$ hence $2y+1=503$, or $y=251$. This gives the solution $(x,y,z)=\boxed{(2,251,252)}$ which can be checked to satisfy the equation. If $y=z$, then $\frac{y+z}{503}=1+\frac1{y^2}$, so $y^2\mid1$. This means that $y=1$ which is impossible as $x\le y$. Case 2: $x=1$ Let $s=y+z$ and $t=yz$. Now we can write the equation as: $$\frac s{2012}=\frac{t+2}{s^2-3t}.$$Note that $t\ge4$, otherwise $(y,z)\in\{(1,2),(1,3)\}$ which is absurd. Using $s^2-3t\ge t$ and $t\ge4$, we have: $$\frac s{2012}\le1+\frac2t\le\frac32,$$so $s\le3018$. Equality holds when $y=z=2$ which is absurd, so this inequality is strict. Using the fact that $503\mid s$, we have that: $$s\in\{503,1006,1509,2012,2515\}.$$From each of these, we can calculate $t$ with the formula:

This gives us no more solutions.

Note $2012=2^2\cdot 503$ and $\gcd(x,xyz+2)=1,2.$ If $503\mid x,$ then $$\nu_{503}(x^3(y^3+z^3))\ge 3>1=\nu_{503}(2012(xyz+2)),$$a contradiction. Similarly, $p\nmid x$ for all primes except $2.$ If $4\mid x,$ then $xyz+2\equiv 2\pmod{4}$ so $$\nu_{2}(x^3(y^3+z^3))\ge 6>3=\nu_2(2012(xyz+1)),$$a contradiction. Hence, $x=1,2$ and $503\mid y^3+z^3.$ Then, $y\equiv(y^3)^{335}\equiv (-z^3)^{335}\equiv -z\pmod{503}$ by FLT so $y+z=503k$ for $k\in\mathbb{N}.$ Case 1: $x=1.$ Factoring yields $k(y^2-yz+z^2)=4(yz+2).$ Notice if $y\neq z\pmod{2},$ then the LHS is odd while the RHS is even. Hence, $2\mid y+z.$ If $k\ge 6,$ then \begin{align*}2(y-z)^2+y^2+z^2-yz-4>3(y-z)^2\ge 0&\implies 3y^2+3z^2-5yz-4>0\\&\implies\text{LHS}\ge 6(y^2-yz+z^2)>4yz+8\end{align*}so $k=2,4.$ If $k=2,$ we know $y^2-yz+z^2=2yz+4,$ or $1006^2=(y+z)^2=5yz+4,$ which is false modulo $5.$ If $k=4,$ then $y^2-yz+z^2=yz+2$ yields $(y-z)^2=2,$ a contradiction. Thus, there are no solutions in this case. Case 2: $x=2.$ Substituting $x=2$ and $y+z=503k$ in the original equation gives $8\cdot 503k(y^2-yz+y^2)=2012\cdot 2(yz+1)$ or $k(y^2-yz+y^2)=yz+1.$ If $k\ge 2,$ then $$(y-z)^2+y^2+z^2-yz-1>2(y-z)^2\ge 0\implies\text{LHS}\ge 2y^2-2yz+2z^2>yz+1,$$so $k=1.$ Then, $y^2-yz+y^2=yz+1$ or $(y-z)^2=1.$ Therefore, $z=y+1$ and $(x,y,z)=(2,251,252).$ $\square$

We claim the answer is $(1,251,252)$ only. Note that $2012=2^2 \cdot 503$, where $503$ is prime. If $503|x$, then we have $$1=v_{503}(xyz+2)+1=v_{503}(2012(xyz+2))=v_{503}(x^3(y^3+z^3)) \ge 3,$$a contradiction. Note that $x|2012(xyz+2) \implies x|4024 \implies x|8$, where the last step is from the above. If $4|x$, then $$6 \le v_2(x^3(y^3+z^3))=v_2(2012(xyz+2))=2+v_2(xyz+2) = 3,$$a contradiction so $x=1$ or $x=2$. We also have $503|y^3+z^3$, so $\left( \frac{-y}{z} \right) ^3 \equiv 1 \pmod{503}$. If the order of $\frac{-y}{z}$ is $3$, then $3|502$, a contradiction. This means $503|y+z$. Case 1: $x=1$. We have $y^3+z^3=2012(yz+2)$, so $2|y+z$. Let $y+z=1006k$. Assume $k \ge 3$. Then $$2(yz+2)=k(y^2-yz+z^2) \ge kyz \implies 1006k-1 \le (k-2)yz \le 4,$$a contradiction. Thus $k=1$ or $k=2$. If $k=2$, we get that $(y-z)^2=2$ a contradiction. If $k=1$, then $y+z=1006$. Note that $$4 \equiv y^2-3yz+z^2 \equiv (y+z)^2-5yz \equiv 1 \pmod{5},$$a contradiction so there are no solutions for this case. Case 2: $x=2$. We have $y^3+z^3=503(yz+1)$. Let $y+z=503k$. If $k \ge 2$, we have $$yz+1=k(y^2-yz+z^2) \ge 2yz,$$a contradiction since $y+z > 2$. This means $y+z=503$, so we have that $(y-z)^2=1$. This means $z=252$ and $y=251$, so we get the solution $(x,y,z)=(1,251,252)$ which is the only solution for this case so we are done.

Clearly $x\mid 2012\cdot 2=503\cdot 2^3.$ Consider $503\mid x$, but then $503^2\mid xyz+2,$ absurd. Hence $x\in \{1,2,4,8\}.$ The largest exponent of $2$ dividing $2012(xyz+2)$ is $2^3,$ we must have $x\in \{1,2\}.$ In either case, $503\mid y^3+z^3$ implies $503\mid y^{501}+z^{501}.$ But by FLT, $503\mid y^{502}-z^{502}$ and so $503\mid y+z.$ If $x=2$, then since $y^2-yz+z^2\geq yz$ we have $503=y+z$ giving solution $(x,y,z)=(2,251,252).$ We refute the case $x=1.$ We have $2\mid y+z.$ Again $y^2-yz+z^2\geq yz$ gives $y+z\in \{1006,2012\}.$ The second has no solution. The former implies $5yz=503^2\cdot 2^2-4$, no solution.

Note $2012=2^4\cdot503$ Claim 1 : $503\mid y+z$ Proof : Let $503\mid x^3 \implies 503\mid x$. By our given condition, $503^3\mid x^3(y^3+z^3)$, but $503^2\nmid 2012(xyz+2)$, so $503\nmid x^3 \implies 503\mid y^3+z^3$. So, $y^{501}\equiv(y^3)^{167}\equiv(-z^3)^{167}\equiv -z^{501}$ and by FLT, $y^{502}\equiv z^{502} $ (mod $503$). So, $y\equiv -z (\mod 503)$, which implies $503\mid y+z$ Claim 2 : $x\in \{1,2\}$ Proof : $x^2(y^3+z^3)=2012yz + \frac {4024}{x}$ and $503\nmid x$, so $x$ is the divisor of $8$. If $x=4,8$, then $x(y^3+z^3)=\frac{2012yz}{x} + \frac {4024}{x^2}$ gives contradiction. So, $x\in \{1,2\}$ Case 1 : $x=1$, No solutions. Proof : $x=1\implies y^3+z^3=2012(yz+2)$. Now, $2012(yz+2)=(y+z)(y^2-yz+z^2)\geq yz(y+z)$ [By AM-GM] $\implies y+z \leq 2012$, since the parity of $y, z$ should be same, Then $y+z=\{1006, 2012\}$. This is easy to find there are no solution in this condition. Case 2 : $x=2$, then $\{y, z\} = \{251, 252\}$. Proof : $x=2\implies (y^3+z^3)=503(yz+1)\implies \frac{y+z}{503}=\frac{yz+1}{y^2-yz+z^2}$, since LHS is a integer and $y^2-yz+z^2\geq yz$, so, $y+z=503$ and $y^2-yz+z^2=yz+1 \implies (y-z)^2=1$. Thus, $\{y, z\} = \{251, 252\}$. So, the solution is $(x, y, z)\equiv (2, 251, 252)$

Note that $2012 = 4 \cdot 503$. Set $p = 503$. Note that if $p \mid x$, then $\nu_p$ of both sides are not equal. Thus $p \mid y^3+z^3$ or $p \mid y+z$ as $p \equiv 2 \pmod 3$. Then $$x^3 \mid 4p(xyz+2) \implies x \mid 8p \implies x \mid 8.$$For $\nu_2$ reasons, $x \in \{1, 2\}$. If $ x = 1$, then $$\frac{y+z}{2p} \cdot (y^2-yz+z^2) = 2(yz+ 2).$$Now we have clear size issues unless $y-z = 0$ or $y+z = 2p$. The former case yields $p^2 \equiv 1 \pmod 5$, and the latter case yields $y^3 = 2p(y^2+2)$, both of which are impossible. If $x=2$, then $\frac{y+z}p \cdot (y^2-yz+z^2) = yz+1$, so again $y^2+yz+z^2 \leq yz + 1$, hence $y=z$ or $y-z = 1$. The former case is impossible, while our latter case yields the only solution set $(2, 251, 252)$ and $(2, 252, 251)$.

Bleh. The same as always. Notice either $503 \mid x^3$ or $503 \mid y^3+z^3$. In the first case we derive contradiction by $v_{503}$. In the second case, since $503 \equiv 2 \pmod 3$, we have that $503 \mid y+z$. So $LHS \geq x^3 \cdot 503 \cdot (y^2-yz+z^2) \geq 503x^3yz$ so since $yz \geq 503$ we have $x \leq 2$. If $x = 2$ this reduces to $(y+z)(y^2-yz+z^2) = 503(yz + 1)$. If $y+z \geq 1006$ this is size contradiction, so $y+z = 503$, then $(y-z)^2 = 1$ so we have $y, z = 251, 252$. Otherwise if $x = 1$, we have $(y+z)(y^2-yz+z^2) = 2012(yz+2)$. By size, we must have $y+z \leq 2012$. So we split into four cases: $y+z = 2012$ fails since $(y-z)^2 = 2$. $y+z = 1509$ fails since $3 \cdot 1509^2 - 13yz = 9$ fails mod $13$. $y+z = 1006$ fails since $1006^2 - 5yz = 4$ fails mod $5$. $y+z = 503$ fails since we have $503^2 - 7y(503-y) = 8$ but this has no solutions in $y$ modulo $5$. Hence the only solution is $(2, 251, 252)$.

Firstly $x\mid 2012(xyz+2)$. If any divisor $d\mid x$ divides $(xyz+2)$, then $d\mid 2$, so $d=1$ or $2$. Since this holds for any divisor, we can only have $x=1,2$ in that case. Otherwise, no divisor of $x$ divides $(xyz+2)$, so $x\mid 2012$, and more generally, $x^3\mid 2012$, but $2012=2^2\cdot 503$ has no cubic divisors, a contradiction. Hence $x\in\{1,2\}$. It is also easy to check that $x$ and $y$ must be both odd. Also clearly, $503$ does not divide both of $y$ and $z$ and also not exactly one of them. Then note that by Fermats little theorem \[y^{502}\equiv z^{502}\mod{503}\]and \[(y^3)^{167}\equiv-(z^3)^{167}\mod{503}.\]Then, dividing the former by the later, we get $y\equiv -z\mod{503}$ or $503\mid y+z$. Thus, let $y+z=503t$. Suppose first that $x=1$. Plugging in, \[t(y^2-yz+z^2)=4(yz+2).\]Note that the RHS seems eventually smaller than the LHS, so we try bounding. First note that $t$ must be even as $y^2-yz+z^2$ is odd. If $t\ge 6$, then \[4yz+8=t(y^2-yz+z^2)\ge 6(y^2+z^2)-6yz\implies 10yz+8\ge 6(y^2+z^2)\ge 12yz\]so $8\ge 2yz$. But as $y+z\ge 503$ and $z\ge y$, we must have $z\ge 252$, making this inequality impossible. Hence $t=4$ or $t=2$. If $t=4$, then \[8yz+8=4y^2+4z^2\implies 2yz+2=y^2+z^2\implies 4yz+2=(2012)^2,\]which is impossible mod $4$. Finally, if $t=2$, then \[1006^2=(x+y)^2=5yz+4,\]but looking mod $5$ gives a contradiction. Hence we must have $x=2$. This gives \[8t(y^2-yz+z^2)=4(2yz+2)=8yz+8\iff t(y^2-yz+z^2)=yz+1.\]If $t\ge 2$, we simply bound again and get a contradiction. Hence $t=1$. Then \[(z-y)^2=1\implies z=y+1\]and so $(2,251,252)$ is the only possibility, and it is indeed a solution.

We claim that the answer is $(x,y,z)=(2,251,252)$. Case 1: $x$ is divisible by 503. Let $x=503c,$ so $$503^3c^3(y^3+z^3)=4\cdot503(503yz+2)$$$$503^2c^3(y^3+z^3)=4(503yz+2),$$but the left side is a multiple of 503 while the right side is 8 mod 503, contradiction. Case 2: $x$ is not divisible by 503. Then, $y^3+z^3$ is divisible by 503. This can be rewritten as $$y^3=(-z)^3\pmod{503}.$$ Claim: Cubes are bijective mod 503. Let $g$ be a primitive root mod 503, and label the nonzero residues $g^0,g^1\cdots g^{501}.$ Thus, since 502 is relatively prime to 3, by Chinese Remainder Theorem, we have that $g^0,g^3,g^6\cdots g^{501\cdot 3}$ covers each of these residues exactly once, hence cubes are bijective mod 503. This means that our previous expression can actually be simplified to just $$y=-z\pmod{503}$$or $y+z\equiv0\pmod{503}.$ Let $y+z=503k$. Then, plugging back in we have $$x^3k(y^2-yz+z^2)=4xyz+8.$$Note that $yz\leq\frac{1}{4}(y+z)^2$, so $$y^2-yz+z^2\geq \frac{1}{4}(y+z)^2.$$This means that $$x^3k(y^2-yz+z^2)\geq \frac{1}{4}x^3k(y+z)^2=\frac{1}{4}503^2x^3k^3.$$Thus, we must have $$4xyz+8\geq \frac{1}{4}503^2x^3k^3$$$$16xyz+32\geq 503^2x^3k^3.$$However, by AM-GM, $$16xyz+32\leq 4x(y+z)^2+32=4\cdot 503^2xk^2+32.$$Combining this with the previous inequality, $$4\cdot 503^2xk^2+32\geq 503^2x^3k^3.$$ Dividing by $503^2xk^2$, this becomes $$4+\frac{32}{503^2xk^2}\geq x^2k.$$Obviously, the left hand side is less than 5, since $\frac{32}{503^2xk^2}\leq \frac{32}{503^2}<1$, so $$x^2k\leq 4,$$as it is an integer anyways. There are only a few more cases to check. Let's go back to the equation $$x^3k(y^2-yz+z^2)=4xyz+8.$$ If $x=1$, then the equation becomes $$ky^2-(k+4)yz+kz^2=8.$$However, we also have $$ky^2+2kyz+kz^2=k(503k)^2=503^2k^3,$$so subtracting these equations gives $$(3k+4)yz=503^2k^3-8,$$so $$yz=\frac{503^2k^3-8}{3k+4}.$$For $k=2,3,4,$ this is not even an integer. For $k=1$, this equals $36143$, so $yz=36143$, but due to $x+y=503$, and $503^2-4\cdot 36143$ not being a perfect square, there are no solutions in this case. If $x=2$, then we must have $k=1$ due to $x^2k\leq 4$, so the equation is $$8(y^2-yz+z^2)=8yz+8$$$$y^2-yz+z^2=yz+1$$$$(y-z)^2=1.$$Since $z\geq y$, we have $z=y+1$, so $y=251,z=252$, which does indeed satisfy the equation, hence we are done.

OTIS Walkthrough Factor $2012 = 2^2 \cdot 503$. Claim: $x \in \{1, 2\}$. Proof. Assume that $503 \mid x$. Clearly then $$\nu_{503}(x^3(y^3 + z^3)) \geq 3$$However as $\nu_{503}(2012(xyz+2)) = 1$ we cannot have $503 \mid x$. Then clearly $x \in \{1, 2\}$ as $\gcd(xyz + 2, x) = \gcd(2, x)$. $\blacksquare$ As a corollary we find $503 \mid y^3 + z^3$. Claim: $503 \mid y + z$ Proof. Note that $y^3 + z^3 = (y+z)(y^2 - yz + z^2)$ is divisible by $503$. Now assume that $503 \nmid y+z$. Then we have $503 \mid y^2 - yz + z^2$. Now clearly $503 \nmid yz$ as then we would have $503 \mid y, z$ which brings a contradiction via $\nu_{503}$. Thus assume that $503 \mid y^2 - yz + z^2$, but $503 \nmid y, z$. Then we also have, $503 \mid 4(y^2 - yz + z^2) = (2y-z^2) + 3z^2$. However then we must have $-3$ as a quadratic residue modulo $503$, contradiction. Thus $503 \mid y + z$. $\blacksquare$ Claim: We must have $\boxed{(x, y, z) = (2, 251, 252)}$. Proof. Now let $y + z = 503k$, allowing us to rearrange the given as, \begin{align*} 4(xyz+2) &= kx^3(y^2 - yz + z^2)\\ 4(xyz+2) &\geq kx^3yz\\ 8 &\geq xyz(kx^2 - 4) \end{align*}Now casework on $k$ and $x$ gives us the claim. $\blacksquare$

Notice $x\mid 4024$ and yet $503\nmid x$ thus $x\in \{1,2,4,8\}$. If $503\mid y^3+z^3$ then \[\left(\frac{y}{-z}\right)^3\equiv 1\pmod {503}\]but since $3\nmid 503-1$ it follows that $y+z\equiv 0\pmod {503}$. Hence write $y+z=503k$. Now \[x^3k(y^2-yz+z^2)=4(xyz+2)\]thus $x\in \{1,2\}$ by simple $\nu_2$ comparison. If $x=1$ then manipulate into \[k(503k)^2=k(y+z)^2=(3k+4)yz+8\le (3k+4)\left(\frac{503k}{2}\right)^2+8\]which yields \[4k\le 3k+4+\frac{32}{(503k)^2}\]which implies $k\le 4$. Note also that \[yz=\frac{k(503k)^2-8}{3k+4}\]and divisibility yields $k=1$ only. Then $yz=36143$. Quadratic formula yields no solutions as the discriminant is actually $7\pmod {10}$. If $x=2$ then repeating logic yields $k=1$ only as the possible solution, which gives $yz=63252$. Then the discriminant of the resulting quadratic is $1$ and thus $y=251$ and $z=252$. The answer is $(2,251,252)$.

Taking mod $x$ on both sides yields $x|4042 = 2^3 \cdot 503.$ Note that $503$ is prime. If $503 | x,$ clearly $503^2$ divides $xyz+2$ which is impossible as $xyz+2 \equiv 2 \pmod {503}.$ Thus $x | 8.$ If $x=4, 8$ observe that $$v_2(\text{RHS}) = v_2(2012)+v_2(xyz+2) = 2+v_2(xyz+2) \leq 3,$$a contradiction. Therefore $x$ is $1$ or $2.$ If $x=2,$ we have $$y^3+z^3=503(yz+1).$$Thus $503|y^3+z^3.$ Letting $a=y/z$ yields $a^3 \equiv -1 \pmod{503},$ so $\text{ord}_{503}(a)=2$ or $6.$ But by FLT $a^{502} \equiv 1 \pmod {503}$ so it follows that $$a^2 \equiv 1 \pmod {503} \implies a \equiv -1 \pmod {503} \implies y+z \equiv 0 \pmod {503}.$$Hence, let $y+z=503k.$ Therefore we have $k(y^2+z^2-yz)=(yz+1) \implies yz+1 \geq y^2+z^2-yz \implies 1 \geq (y-z)^2.$ If $y=z$ then $503|y, z$ which yields a contradiction by considering $v_{503}$ of both sides. WLOG, assume that $y > z,$ then $y=z+1.$ After solving the resultant equation, we get $(x, y, z) = (2, 251, 252)$ and $(2, 252, 251).$ Meanwhile, if $x=1$ we have $2012(yz+2) = (y+z)(y^2+z^2-yz).$ But it is easy to see $y, z$ have the same parity and $y \neq z$ so if for the sake of a contradiction $y+z \geq 2012$ we have $2012(yz+2) = (y+z)(y^2+z^2-yz) \implies 2 \geq (y-z)^2$ which is a contradiction. Therefore $y+z < 2012,$ but using similar logic as above we can deduce that $503|(y+z).$ Therefore $y+z=503, 1006, 1509.$ If $y+z=503$ or $1006,$ both these fail mod $5.$ If $y+z=1509,$ we have that $6831234=13yz$ but the LHS isn't divisible by $13,$ a contradiction. Thus there are no solutions in this case. Therefore, the only solutions are $(2, 251, 252)$ and $(2, 252, 251).$

Taking mod $x$ on both sides yields $x|4042 = 2^3 \cdot 503.$ Note that $503$ is prime. If $503 | x,$ clearly $503^2$ divides $xyz+2$ which is impossible as $xyz+2 \equiv 2 \pmod {503}.$ Thus $x | 8.$ If $x=4, 8$ observe that $$v_2(\text{RHS}) = v_2(2012)+v_2(xyz+2) = 2+v_2(xyz+2) \leq 3,$$a contradiction. Therefore $x$ is $1$ or $2.$ If $x=2,$ we have $$y^3+z^3=503(yz+1).$$Thus $503|y^3+z^3.$ Letting $a=y/z$ yields $a^3 \equiv -1 \pmod{503},$ so $\text{ord}_{503}(a)=2$ or $6.$ But by FLT $a^{502} \equiv 1 \pmod {503}$ so it follows that $$a^2 \equiv 1 \pmod {503} \implies a \equiv -1 \pmod {503} \implies y+z \equiv 0 \pmod {503}.$$Hence, let $y+z=503k.$ Therefore we have $k(y^2+z^2-yz)=(yz+1) \implies yz+1 \geq y^2+z^2-yz \implies 1 \geq (y-z)^2.$ If $y=z$ then $503|y, z$ which yields a contradiction by considering $v_{503}$ of both sides. Therefore $z=y+1.$ After solving the resultant equation, we get $(x, y, z) = (2, 251, 252).$ Meanwhile, if $x=1$ we have $2012(yz+2) = (y+z)(y^2+z^2-yz).$ But it is easy to see $y, z$ have the same parity and $y \neq z$ so if for the sake of a contradiction $y+z \geq 2012$ we have $2012(yz+2) = (y+z)(y^2+z^2-yz) \implies 2 \geq (y-z)^2$ which is a contradiction. Therefore $y+z < 2012,$ but using similar logic as above we can deduce that $503|(y+z).$ Therefore $y+z=503, 1006, 1509.$ If $y+z=503$ or $1006,$ both these fail mod $5.$ If $y+z=1509,$ we have that $6831234=13yz$ but the LHS isn't divisible by $13,$ a contradiction. Thus there are no solutions in this case. Therefore, the only solution is $(2, 251, 252).$