Solution for December 2012
The Problem: 

Find all positive integers $n$ such that $n+184$ and $n285$ are both cubes of integers.

The solution:
Correct solutions were submitted by
Bahman Ahmadi (Regina) 
Lamis Alsheikh (Syria) 
Diana Andrei (Sweden) 
Arkady Arkhangorodsky (Ontario) 
Claudio Baiocchi (Italy) 
Lou Cairoli (USA) 
Bernard Carpentier (France) 
Ruben Victor Cohen (Argentina) 
Bernard Collignon (France) 
Hubert Desprez (France) 
MeiHui Fang (Austria) 
Federico Foieri (Argentina) 
Philippe Fondanaiche (France) 
Georges Ghosn (Montréal) 
Kerim Gokarslan (Turkey) 
Tony Harrison (England) 
Seongwoo Hong (USA) 
Codreanu Ioan (Romania) 
Alex Jeon (USA) 
Kipp Johnson (USA) 
Normand LaLiberté (Ontario) 
Matthew Lim 
Patrick J. LoPresti (USA) 
Sanjeev Nimishakavi and & Milan Pavić (Serbia) 
Nawal Kishor Mishra (India) 
Mathias Schenker (Switzerland) 
Armend Sh. Shabani (Kosovo) 
Albert Stadler (Switzerland) 
Andreas Stahl (Germany) 
Hakan Summakoğlu (Turkey) 
Bruno Tisserand (France) 
Daniel Văcaru (Romania) 
Arthur Vause (England) 
Paul Voyer (France) 
Wu ChengYuan (Singapore) 

In addition, we received one incomplete solution.
We present a composite of the best ideas from all the submitted solutions. Let
$$a^3 = n+184 \quad \mbox{ and } \quad b^3 = n285 .$$
Then $a^3b^3 = (n+184)  (n285) = 469$.
Factoring the left and right sides gives us
$$(ab)(a^2+ab+b^2) = 7 \cdot 67.$$
Because $a^2+ab+b^2 \ge 0$ for any real numbers $a$ and $b$, it follows that $ab > 0$. Furthermore, because $a$ and $b$ are integers, $a^2+ab+b^2 > (ab)^2 \geq ab > 0$, whence the only possible values for $ab$ are the factors 1 and 7 (so that $a^2+ab+b^2$ takes on the respective values $\frac{469}1=469$ and $\frac{469}7= 67$). To find $a$ and $b$ in each of these cases, note that $a^2+ab+b^2  (ab)^2 = 3ab$. For $ab = 7$ we have
$$a^2+ab+b^2 =67, \quad 3ab = 677^2 = 18,$$ $$ \quad (a+b)^2 = a^2+ab+b^2 + ab = 67+\frac{18}{3} = 73.$$
But 73 is not a perfect square, which means that $ab \ne 7$.
For $ab=1$ we have
$$a^2+ab+b^2 =469, \quad 3ab = 468, \quad (a+b)^2 = a^2+ab+b^2 + ab = 469+\frac{468}{3} = 625.$$
Thus, we must have $a+b = \pm25$, and either
$$25+1 = (a+b)+(ab) =2a, \; \mbox{ whence } \; a=13, \; b=12,$$
or
$$25+1 = (a+b)(ab) =2a, \; \mbox{ whence } \; a=12, \; b=13.$$
In the first case $n = a^3184 =2197184 = 2013 \;(= 12^3 + 285)$; in the second case,
$n = a^3184 =1728184 = 1912 \; (= 13^2 + 285)$. Thus, the only possible positive value of $n$ is 2013, as claimed.
Comments. Our December problem was suggested by the Correspondence Mathematical Competition in Slovakia 2006/7, first round, first set, problem 6 (as reported in Crux Mathematicorum with Mathematical Mayhem, 37:8 (December 2011) pages 507508. Kipp Johnson observed that if we assign the value $9x+50$ to the letter of the alphabet in position $x$ (so, for example, $a$ in position 1 is assigned 59, $b$ gets 68, and so on), the sum of the values of all the letters in the phrase Happy New Year is 2013. We are embarrassed to admit that we overlooked this obvious connection to our problem. Nevertheless, we wish you all a healthy and prosperous 2013 in which all your problems have easy solutions.
