My sketch: a) Let b = c = d = 1 and a = 3k + 1. Then n = 9k^3 + 9k^2 + 3k + 1 is special for any positive integer k. Clearly, these numbers are infinitely many. b) 2014(c^3 + 2d^3) = a^3 + 2b^3. Since gcd(3, 19 - 1) > 1, it's good to consider mod 19, because cubes don't form a complete residue system mod 19. Then, we easily get a contradiction, since the numbers are divisible by 19^m, for every m.

Indeed. Point a) is ridiculous by its triviality; taking arbitrarily $c,d,k \in \mathbb{N}^*$ and $a=kc$, $b=kd$, we get $\dfrac{a^3+2b^3}{c^3+2d^3} = k^3$, thus all not-null perfect cubes $n=k^3$ are special. Similarly trivial is to take $c=d=1$ and $a\equiv b \pmod{3}$. i) Equality $a^3+2b^3 = 2014(c^3+2d^3) = 2\cdot 19\cdot 53(c^3+2d^3)$ cannot take place. From $19\mid x^3+2y^3$ follows $x\equiv y \equiv 0 \pmod{19}$, since cubic residues modulo $19$ are $\{0, \pm 1, \pm 7, \pm 8\}$, thus $-2$ is not a cubic residue modulo $19$. But then it follows $a\equiv b \equiv 0 \pmod{19}$, hence $c\equiv d \equiv 0 \pmod{19}$, and an infinite descent argument raises the contradiction.

This problem was submitted by Romania. Does anyone know who proposed it?

Oh, I wouldn't even want to know that Such stones are better left unturned (meaning the problem is so trite, its author should beg for anonimity). By the way, any info on which country proposed the other (three) problems? Maybe Serbia for Problem 4?

mavropnevma wrote: Oh, I wouldn't even want to know that Such stones are better left unturned (meaning the problem is so trite, its author should beg for anonimity). By the way, any info on which country proposed the other (three) problems? Maybe Serbia for Problem 4? Haha... I would want to know regardless Problems 1 and 4 by the UK (David Monk and Sahl Khan, respectively), and Problem 3 by Greece.

All the problems very easy indeed. The geometry could be done from many approaches. And the combo wasn't that hard if you had enough time on it after you had solved the first 3 problems.

absolutely, you are right, @ifetahu!

B) a^3+2b^3=2*19*53k:19=> a^18=2^6*(b^18)(mod 19) by little fermat we have a^18=1(mod 19) => if (a,19)=1 we have 2^6-1:19 we false the we have a,b:19 => alogrlously we have that c^3+2d^3:19 so we can do this infinetly we false and we have contradiction

mathworld1 wrote: This problem was submitted by Romania. Does anyone know who proposed it? Either Mihai Baluna, or Dan Schwarz, or Bogdan Enescu, or Radu Gologan, or Dinu Serbanescu?

Oh, cross out the second option "Nevah", as Churchill would have uttered it (never). I can also offer to cross out the fifth (last) one. As for the others ... all things are possible upon the earth, could have said J. L. Borges.

Gherghe Cătălin-Liviu was the team leader for Romania, Marius Mâinea the deputy team leader, and Radu Gologan Observer A. Perhaps one of these three?

Only viable possibility - number three (overlapping with previous guesses). Number two - a definite impossibility There is a secretomania prevalent in Romania (justified in this case, by my previous remark, ha ha).

mavropnevma wrote: Only viable possibility - number three (overlapping with previous guesses). Number two - a definite impossibility There is a secretomania prevalent in Romania (justified in this case, by my previous remark, ha ha). What secretomania? By the way, maybe the author is hiding him/herself because of your previous remark: "Such stones are better left unturned (meaning the problem is so trite, its author should beg for anonimity)."

Why should there even be part (a)? To feed free points to everyone? (or part (b) seems to "wimpy" by itself, so adding part (a) is just a desperate attempt to cover up its "wimpiness", but just made it worse). Well, I got "trolled" for about 5 minutes in (a) before realizing the stuff ... Well, maybe they didn't notice that the $19$-th power cubic residues could be listed easily enough, and the team leaders thought that FLT is needed. Honestly, I think that this paper is suited more for JBMO than BMO ... Does anyone know how many points are (a) worth and (b) worth?

Quote: Honesly, I think that this paper is suited more for JBMO than BMO... I totally agree.I think that the inequality was too easy even for a JBMO Problem 1

Konigsberg wrote: Does anyone know how many points are (a) worth and (b) worth? a) worth 2 points, while b) worth 8 points. Same as Konigsberg I was "fooled" by a) and first started to work on b), because I thought it would be easier to work with "fixed number" n = 2014. Immediately I started working modulo 19 and unlike the official solution, which uses FLT, I listed all cubic residues modulo 19 and finished the proof as mentioned in the official solution. a) should be obvious by plugging some "small" integers and noticing $c=d=1$ and $a=b$ provide that every cube is a "special" number. One reason why they asked easier problems this year is maybe the fact that the previous year the problems were much more difficult, and medal boundaries were relatively low, compared to past years. Also, the official solution for Problem 1 is over-complicated. I can't remember exactly, but I think it uses AM-GM and C-S, while I and most of the participants solved it using simply AM-GM.

Is my solution on the Problem 1 correct?

Yeah this problem was also very easy.The first part follows directly from the following fact: $3^{3k}=\frac{(3^{k+1})^3+2 \times (3^{k+1})^3}{3^3+2 \times 3^3}$ I did the second one similar to the others in this forum.(and I think this is the only way).

each number in a form of $ x^3 $ is special: \[ x^3 = \frac{x^3 + 2*x^3 }{1^3 + 2 * 1^ 3} \]

if we take $c,d=1$ and $a=3k+1$ $b=3t+1$ we get that $a^3+2b^3\equiv 0mod3$

My solution. $a)$ Let $a=kc$ and $b=kd,$ then every perfect cube be special number. $b)$ $2\cdot 19\cdot 53(x^3+2y^3)=a^3+2b^3.$ Let $a,b,x,y$ be the minimal value such that satisfies the condition of $b.$ We know $19\mid a^3+2b^3\to 19\mid a,b.$ If $a^3\equiv -2b^3\mod 19\to a^{18}\equiv 2^6b^{18}\mod 19\to 2^6\equiv 1\mod 19.$ But this is impossible. Then $a=19a_1,b=19b_1.$ But this is impossible for our condition (minimal condition). So we are done.

Part A: set a=b=c=d to get that n=1 is achievable. Then multiply a and b by some k gives that k^3 for all integers k are all achievable, that is indeed infinite. Part B: Use a modulus argument. Take mod 19, we show that it is not possible for this to be true. Bashing through all the third powers modulus 19 we derive that a^3 + 2b^3 is not possible to be 0 mod 19 unless both are 0 mod 19. Since 2014 only has 1 multiple of 19, we must have the denominator cancel out at least 19^2. But this means denominator is divisible by 19 which by the previous result means it is divisible by 19^3, forming an infinite loop. Thus we can assume that a, b are relatively prime mod 19, which does NOT provide a valid result. Thus 2014 cannot be formed.

i) $c=d=1$ and $a=b=3k$ gives infinitely many n ii) Assume 2014 is a special number $2014=\frac{a^3+2b^3}{c^3+2d^3}\iff a^3+2b^3=19\cdot106(c^3+2d^3)$ $a^3\equiv1,7,8,11,12,18\pmod{19}$ $2b^3\equiv2,3,5,14,16,17\pmod{19}$ Combining both results we get that $a^3+2b^3\not\equiv0\pmod{19}$ contradiction $\implies$ 2014 isn't special