# find all triples $(x,y,z)$ of positive integers such that $x\leq y\leq z$ and $x^3 (y^3 + z^3)=2012(xyz+2)$

(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 $$y,z\not\equiv 0\mod 503.$$

• Is there a question in here?
– Mike
Commented Nov 29, 2022 at 17:09
• @Mike I've updated my question. Commented Nov 29, 2022 at 17:21

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.