Find all $ (x,y,z) \in \mathbb{Z}^3$ such that $ x^3 - 5x = 1728^{y}\cdot 1733^z - 17$. (Paolo Leonetti)
Problem
Source: Own (Oliforum contest 2009/round2/3)
Tags: modular arithmetic, calculus, integration, algebra, polynomial, function, quadratics
Zhero
19.10.2009 02:31
Rearranging, we have $ x^3 - 5x + 17 = 1728^y 1733^z$. Since 1733 and 1728 are relatively prime, we must have $ y \geq 0$ and $ z \geq 0$, or else the right-hand side will not be an inteegr. Since $ x^3 - 5x + 17$ is always odd, we must have $ y \leq 0$, so $ y = 0$ (otherwise, the right-hand side would be even.) Thus, we seek all integers $ x$ and nonnegative $ z$ such that $ x^3 - 5x + 17 = 1733^z$.
Suppose $ z > 0$. Then we have that there exists some solution $ x$ to $ x^3 - 5x + 17 \equiv 0 \pmod 1733$. But no such solution exists, so $ z = 0$. However, the rational root theorem implies that this has no integral solutions, so we get that there are $ \boxed{\mbox{no solutions}}$.
On the actual contest, I did not provide a legitimate proof (meaning I printed all 1733 possible outputs of $ x^3 - 5x + 17$ into a file and copy/pasted it into my document) that the polynomial has no zeroes mod 1733. So that's a hole in the proof, I suppose. I have an inkling of a feeling that it follows from the fact that the term under the square root here, that is, 197181, is not a quadratic residue mod 1733. In any case, if I'm not mistaken, rigorizing this argument will involve working in field extensions, which I'm not very familiar with, and which certainly does not qualify as elementary (as the rules asked for an elementary solution.)
I tried taking the equation $ x^3 - 5x + 17 = 1733^z$ random mods as well, and that didn't turn out so well. There's probably a better solution along that route.
azjps
19.10.2009 05:04
$ \mod{7}$ gives that $ z \equiv 0 \pmod{3}$, and then some bounding upon $ x > 1733^z$, \[ x^3 - (1733^k)^3 = (x - 1733^k)(x^2 + 1733^kx + 1733^{2k}) = 5x - 17\] wins. Though I computed all residues $ \mod{1733}$ also, for good measure [and lack of time ].
Zhero
19.10.2009 06:00
FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL.
Darn. Oh well. That is pretty nice.
mavropnevma
19.10.2009 09:40
Answer to a question in Zhero's post. Imposing the perfect square condition on the discriminant of the depressed cubic $ x^2-5x+17$, as given by Cardano's formula, does not work. There are primes $ p$ for which that value is not a square residue, and still the cubic does have a root modulo $ p$, so it's not a necessary criterium (it is of course sufficient, since it readily yields a root).
bboypa
19.10.2009 13:24
Zhero wrote:
FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL. FAIL.
Darn. Oh well. That is pretty nice. LOL thanks My question is, since this question is raised: there exist some method to see if a cubic polynomial is irriducible in $ F_p$? It seems not a trivial one:is it what you are referring mavropnevma? I am interested in, it can be a very powerful tool (infact it has proved that $ x^3-5x+17=1733y$ is impossible in $ \mathbb{Z}^2$ )
shoki
19.10.2009 16:40
for the case $ z=3k$ we can put the expression between two squares,namely, $ x^3$ and $ (x-1)^3$. and we r done!
bboypa
19.10.2009 20:09
$ 2 \nmid f(x): =x^3-5x+17 \in \mathbb{Z} \implies \min\{y,z\} \in \mathbb{N}\text{ and } y=0$. If $ z=0$ there are no solutions, so $ f(x) \ge 1733 \implies x>10$ Now $ f(x) \in \{0,1,3,5,6\}$ in $ \mathbb{F}_7$ so we have the only case $ 3 \mid z$. So $ f(x)$ must be a cube, but $ (x-1)^3<f(x)<x^3$ for all $ x>10$. Remark: 7 is not a magic conguence since 2.3+1=7, and $ x^3=\left(\frac{x}{7}\right)$ in $ \mathbb{F}_7$. Now $ 1733^z=4^z \in \{1,2,4\}$. 7 does not divide f(-1)f(0)f(1) and if f(x)=2 then $ x^3-5x+17=3 \pm 1 \implies$ $ \left(\left(\frac{x}{7}\right)\pm 1\right)+2x=0$: both addends are odd, and the first one has absolute value 2, so |x|=1, then is a contradiction.
darknmt
08.01.2010 15:43
azjps wrote: \[ x^3 - (1733^k)^3 = (x - 1733^k)(x^2 + 1733^kx + 1733^{2k}) = 5x - 17\] It's seem to be easy to check there is NO solutions $ x$ that $ x < 4$ If $ x > 3$, It's obvious that $ (x - 1733^k)(x^2 + 1733^kx + 1733^{2k}) = 5x - 17 > 0 \rightarrow x > 1733^k$ so $ 5x - 17 > x^2 + 1733x + 1733$ (Clearly no solutions)