Solution for September 2011
The Problem: 

Find the positive integers $m$ and $n$ for which the roots of the equations
$$x^2  mx + (n+1) = 0 \quad \mbox{ and } \quad x^2  (n+1)x + m = 0$$
are positive integers that, together with $m$ and $n$, form in some order an arithmetic progression whose sum is $21$.

The solution:
The September problem came from an Olympiad training problem set compiled by Ed Barbeau in 1994. Correct solutions were submitted by
Jose Arraiz (Brasil) 
Diana Andrei (Sweden) 
Quentin Baudenon (France) 
Luigi Bernardini (Italy) 
Lou Cairoli (USA) 
Bernard Collignon (France) 
Ignacio Somma Esteves (Uruguay) 
MeiHui Fang (Austria) 
Philippe Fondanaiche (France) 
Jan Fricke (Germany) 
Wilfrid Gao (USA) 
Gruian Cornel (Romania) 
Benoît Humbert (France) 
Tom Holens (Manitoba) 
Ile Ilijevski (Macedonia) 
Kipp Johnson (USA) 
Omran Kouba (Syria) 
Normand Laliberté (Ontario) 
Lethe Li (USA) 
Roopesh Mangal (Malaysia) 
Remo Mantovanelli (Italy) 
Omran Kouba (Syria) 
Alex Love (Ontario) 
Christian Pont (France) 
Nawal Kishor Mishra (India) 
Mathias Schenker (Switzerland) 
Albert Stadler (Switzerland) 
Hakan Summakoğlu (Turkey) 
A. Teitelman (Israel) 
Vijaya Prasad Nalluri (India) 
Paul Voyer (France) 
Matthew Lim (USA)

The first step in all the submitted solutions was to observe that the six integers must be distinct (since a trivial arithmetic progression of the same integer repeated six times cannot have a sum of 21), and the smallest possible arithmetic progression of six distinct positive integers sums to 21; consequently,
the four roots and the two parameters $m$ and $n$ are the numbers 1 through 6 in some order.
(In fact, the information that the numbers form an arithmetic progression is redundant: any set of six distinct positive integers has a sum of at least 21.)
Our solvers devised four ways to finish the solution. Let us denote the roots of the first quadratic by $r_1,s_1$, and of the second by $r_2, s_2$.
Method 1. Nearly all solvers observed that $m+n=10$; here is their argument. We are given that $$r_1+s_1+r_2+ s_2+m+n=21,$$
and we know that the roots of a monic quadratic equation add up the the negative of the coefficient of the $x$term:
$$r_1+s_1 = m \quad \mbox{ and } \quad r_2+ s_2 = n+1.$$
Put these equations together to get
$$21 = (r_1+s_1)+(r_2+ s_2)+m+n = m+n+1+m+n = 2m + 2n +1,$$
whence $m+n=10$, as claimed. Next, because $m$ and $n$ are at most 6 and distinct, the only possibilities are $m=6$ and $n=4$, or $m=4$ and $n=6$. The latter pair produces a pair of quadratics that fail to have integer solutions. With $m=6$ and $n=4$, however, the quadratics become
$$x^2  6x + 5 = 0 \quad \mbox{ and } \quad x^2  5x + 6 = 0,$$
whose solutions are 1 and 5 (for the first) and 2 and 3 (for the second), satisfying all our requirements.
Method 2. Here we again use the fact that the sum of the roots of each quadratic equation equals the negative of the coefficient of the $x$term and, what's more, their product equals the constant term. Because we are dealing with positive integers that satisfy
$$r_1+s_1=m=r_2s_2, \quad \mbox{ and } \quad r_2+s_2 = n+1 = r_1s_1,$$
we deduce that $m$ and $n+1$ are at least as large as any of the roots. That means $m$ and $n$ can be only 4, 5, or 6. We know from step 1, moreover, that one of the roots must be 1, and that 1 could not be a root of equation 2 because we would then have $1 \cdot s_2 = m$, contradicting the requirement that $m$ be distinct from $s_2$. That leaves only the possibility that $1 \cdot s_2 = n+1$, whence $s_2 = 5$ (it cannot equal 6 because $m > s_2$) and therefore $n=4$.
Method 3. Here we again use the fact that $r_1s_1 = n+1$ and $r_2s_2 = m$. The latter tells us that the product of two of the numbers from 1 to 6 equals the third  the only possibility is $2 \cdot 3 = 6$, which gives us right away that $m=6$. We are left with 1, 4, and 5 to arrange in the equation $r_1s_1 = n+1$, which forces $n$ to be 4.
Method 4. The discriminants of the given quadratics, namely
$$m^24(n+1) \quad \mbox{ and } \quad (n+1)^2  4m, $$
must be perfect squares. By checking all 30 pairs $m$ and $n$ of distinct integers from 1 to 6, one finds that only the choice $n=4$ and $m=6$ produces a perfect square in the second discriminant. Happily, it produces a perfect square also in the first discriminant.
