(USA Winter TST 2013, December 13 2012) Find all triples $(x,y,z)$ of positive integers such that $x\leq y\leq z$ and $x^3 (y^3 + z^3)=2012(xyz+2)$.
The solution below is far from complete; it doesn't confirm whether there are any solutions to the equation.
For a prime p and positive integer n, let $v_p(n)$ denote the highest exponent of p dividing $n$. There are various properties of $v_p$ such as $v_p(n!) = \sum_{i=1}^\infty \lfloor \dfrac{n}{p^i}\rfloor , v_p(a+b) \ge \min\{v_p(a),v_p(b)\}$ with equality iff $v_p(a)\neq v_p(b)$, $v_p(ab) = v_p(a) + v_p(b),$ etc. Also let $L$ denote the LHS of the equation and $R$ denote the RHS. Since $x\leq y\leq z,$ we must have $2x^6\leq x^3(y^3 + z^3) \leq 2z^6.$ Also, $x^3 \leq xyz\leq z^3$. It might be possible to find an upper bound on $z$. $2012 = 4\cdot 503$ and $503$ is prime. Hence $503$ either divides $x^3$ or $y^3 + z^3 = (y+z)(y^2 – yz+z^2).$ Also, if $x$ is odd, then $y$ and $z$ must have the same parity. Suppose first that $503$ divides $x^3$. Then $503^3$ divides $x^3$ and so $xyz+2$ must be divisible by $503^2$. But this is impossible since $503 | x$ implies that $xyz + 2$ is coprime to $503$. Hence we must have that $503$ divides $y^3 + z^3$.
Now if $x$ is odd, then $xyz+2$ is coprime to $x$. But we've already deduced that $x$ cannot be divisible by $503$ above, so if $x$ has an odd prime factor p, then $p\neq 503$. As well, $p$ does not divide $2012(xyz+2)$ since it divides none of $503, 4, xyz+2$. Hence we get a contradiction if $x$ has an odd prime factor. Thus, we know that $x$ has no odd prime factors, which means that it must be a power of $2$. Henceforth write $x = 2^k$ for some $k\ge 0$. Now before we move on to consider $y$ and $z$, first note that $k$ cannot exceed $1$. For if this were the case, then $v_p(x^3(y^3+z^3)) \ge 6$ while $v_p(2012 (xyz+2)) = v_p(2012) + v_p(xyz+2) =3$, giving a contradiction.
We have thus shown the following claim, which we'll call claim 1: Hence $x=1$ or $x=2$ are the only possibilities.
Claim 2: $\gcd(y,z)$ is a power of $2$.
Suppose $p$ is an odd prime factor that divides $y$ and $z$. Then $p^3$ divides $y^3 + z^3$ so $L=R$ implies $xyz + 2$ must be divisible by $p^2$, which is impossible as $xyz+2$ is coprime to $p.$ Thus $\gcd(y,z)$ is a power of $2$.
Claim 3: $y > 1$. If $y = 1,$ then $x^3(z^3+1) = 2012(xz + 2).$ If $x=1,$ then $z^3 + 1 = 2012(z+2)$. By the Rational Roots Theorem, The positive integer roots of the cubic equation can only be a divisor of $4024$. $z$ clearly exceeds $8$. So $z$ must be divisible by $503$, implying that $2012(z+2) = z^3+1$ is coprime to $503$, a contradiction. If $x=2, (z^3 + 1) = 503(z+1)\Rightarrow z^2 – z + 1 = 503\Rightarrow z^2 – z-502 = 0,$ which has no integer solutions as $1+4\cdot 502 = 2009$ is not a perfect square. Hence $y > 1$.
So $503$ divides $y+z$ or $503$ divides $y^2 – yz+z^2$. Suppose $503$ divides $y+z$. Then $y$ and $z$ must both be coprime to $503$ by claim 2 (indeed if $503$ divides either of $y$ or $z$, then it divides the other, contradicting $503\nmid \gcd(y,z)$). Hence $y$ and $z$ are both coprime to $503$ and $y\equiv -z \mod 503.$ Then $y^2 – yz + z^2\equiv 3z^2 \mod 503,$ so $y^2-yz+z^2$ is coprime to $503$ and the exponent of $503$ on the LHS of the equation is equal to $v_{503}(y+z)$. Suppose $v_{503}(y+z)\ge 2$. Then $xyz+ 2$ must be divisible by $503$, so $xz^2 \equiv 2\mod 503\Rightarrow x\equiv 2z^{-2}\mod 503.$ If $x=1$, then $2z^{-2} \equiv 1\mod 503$. Assume $x=2$. then $z^{-2}\equiv 1\mod 503\Rightarrow z^2\equiv 1\mod 503\Rightarrow z\equiv \pm 1\mod 503.$ WLOG (the other case is similar), assume $z\equiv 1\mod 503.$ Then $y\equiv -1\mod 503$ and $$.
If $yz+2$ is divisible by $503^2,$ then $yz$ is at least $503^2-2$. So
$y^3 + z^3 \ge 2\sqrt{y^3 z^3}$ by the AM-GM inequality, and equality only holds if $y=z$ and both are perfect squares. This is possible only if $y$ and $z$ are even powers of $2$ by claim 2. Now $2\sqrt{yz} (yz) > 2012(yz+2)\Leftrightarrow \sqrt{yz}(yz) > 1006(yz+2)$. So if $yz > 1007^2$, then $\sqrt{yz} > 1007$ and so $\sqrt{yz}(yz) > 1007(yz) > 1006(yz+2).$ Hence we must have $yz\leq 1007^2$ (if $x=2$ and $yz > 1007^2$ then $x^3(y^3 + z^3 ) \ge 16 \sqrt{y^3z^3} > 8 (2012(yz+2)) > 2012(2yz+2)$). I'm not sure if this bound can be strengthened to something more useful. Perhaps the power-mean inequality or the Cauchy-Schwarz inequality could be useful? We have $(y^3 + z^3)/2 \ge (y^{3/2}+z^{3/2})$ by the power mean inequality.
Now suppose $503 | (y^2 – yz+z^2) = (y-z)^2 + yz.$ Then $-yz\equiv (y-z)^2\mod 503.$ As before, we must have $y,z\not\equiv 0\mod 503.$
Best Answer
First, observe that we should have $x\le12$. Otherwise:
$$x^3(y^3+z^3)\ge x^6+x^3(xyz)\ge 13^6+2197(xyz)\gt2012(xyz+2).$$
Therefor $x \in \{1,2,3,4,5,6,7,8,9,10,11,12\}$.
But $x$ cannot be $3, 5,6 , 7, 9, 10, 11$ or $12$ because:
$$x|2012(xyz+2) \implies x|2012 \times2 \implies x|8 \times 503.$$
On the other hand $x$ cannot be $4$ or $8$ either because, in this case, we should have:
$$4^3|4 \times 503(4yz+2) \ or \ {} 8^3|4 \times 503(8yz+2),$$
both of which are impossible.
Therefore $x$ is either $1$ or $2$, and we have two new equations:
$$y^3+z^3=2012(yz+2) \\ y^3+z^3=503(yz+1).$$
Now, note that if $503|y$ then we get $503|z$, hence $503|y+z.$
Let's assume $503 \not| \ y$. Hence $503 \not| \ z$. By the Fermat's theorem, we have:
$$503|z^{502} - y^{502};$$
and because of $503| y^3+z^3$ we conclude that $503| y^{501}+z^{501}$, and as a result $503|y^{502}+yz^{501}$. Therefore: $$503|z^{502}+yz^{501}\implies 503|y+z \implies y+z=503k.$$
Now, we have two other new equations: $$ (1) \ k(y-z)^2+(k-4)yz=8 \\ (2)\ k(y-z)^2+(k-1)yz=1.$$
In the first case, it is very easy to see that the equation has no solution. First notice that $k$ should be $2$, and then we have $(y+z)^2-5yz= 4$ while $y+z=503 \times 2$. This means that the equation has no solution.
For the second equation, it is clear that $k$ must be equal to either $1$ or $2$. If $k=2$ then $z=y=1$, which is impossible. Thus $k=1$, and the only solution is $(2,251,252).$
EDIT: This problem is a shortlist problem of IMO $2012$ and has an official solution.