Solution for March 2010
The Problem: 

Here is a 3part problem for month number 3:
 Can you find integers s and t, both greater than 1, for which (s^{s})^{2} = t^{t}?
 Can you find integers s and t, both greater than 1, for which ( s^{s})^{2010} = t^{t}?
 Can you find integers s and t, both greater than 1, for which s^{(s2010)} = t^{t}?

Correct responses: 

The source for our March problem was Stan Wagon's "Problem of the Week" number 1038 (from September 26, 2005). Solutions were submitted to us by
Bojan Basic (Serbia) 
Magnus Jakobsson (Sweden) 
Shai Covo (Israel) 
Wolfgang Kais (Germany) 
MeiHui Fang (Austria) 
Matthew Lim (USA) 
Philippe Fondanaiche (France) 
John T. Robinson (USA) 
Cornel Gruian (Romania) 
Jan van Delden (The Netherlands) 
Benoît Humbert 
Ray Van Raamsdonk 
A few of these solutions contained unsubstantiated claims that led to faulty conclusions. Because their arguments were essentially correct, we have included the authors' names in the above solver list, but we remind everybody that in mathematics one must make certain that every claim is correct and backed by a valid argument.

The solution:
Part a. No, there are no integers s and t greater than 1 for which (s^{s})^{2} = t^{t}.
Note first that (s^{s})^{2} = s^{2s}. Should there exist such a pair s and t, then on the one hand because s,t >1 we get s^{s} < s^{2s} = t^{t}, whence s < t; on the other hand, t^{t} = s^{2s} < (2s)^{(2s)}, so that t < 2s. But t < 2s implies that s^{t} divides evenly into s^{2s} = t^{t}, whence s would necessarily divide evenly into t; however s < t < 2s tells us that to the contrary, the smallest multiple of s is greater than t, so that s cannot divide evenly into t. This contradiction implies the nonexistence of our desired pair of integers.
Part b. Yes, there are many possible pairs; for example,

and 
s = 2009^{2009} and t = 2009^{2010}. 

To discover the first pair we have
(s^{s})^{2010} = s^{2010s} = s^{2·1005s} = (s^{2})^{1005s},
which is in the form t^{t} if we set s^{2} = 1005s; that is, s = 1005.
Comment. Note that this trick works for all even exponents except 2:
for all integers k > 1, (s^{s})^{2k} = t^{t} when s = k and t = k^{2}. 
Try it with k = s = 2 and t = 4: (2^{2})^{4} = 2^{8} = 4^{4}.
Later we will see where the second pair comes from. For now let us be satisfied in observing that
for all integers n > 2, (s^{s})^{n} = t^{t} when s = (n – 1)^{n–1} and t = (n – 1)^{n}. 
Thus, in the smallest case n = 3 we have s = 2^{2} = 4, t = 2^{3} = 8, and
(s^{s})^{3} = (2^{2})^{3·22} = (2^{3})^{2·22} = (2^{3})^{(23)}; that is, 4^{12} = 8^{8}.
For our problem n = 2010, s = 2009^{2009}, and t = 2009^{2010}:
(s^{s})^{2010} = (2009^{2009})^{2010·20092009} = (2009^{2010})^{2009·20092009}
= (2009^{2010})^{(20092010)} = t^{t}.
In general,
(s^{s})^{n} = ((n1)^{n1})^{n·(n1)n1} = ((n1)^{n})^{(n1)·(n1)n1} = ((n1)^{n})^{((n1)n)} = t^{t}.
Comment. We have here what might be called an antiFermat theorem: It has no solution in integers when the exponent is 2, yet many solutions when the exponent is an integer greater than 2; moreover, the proof would fit in the margin of Fermat's book.
Part c. Yes, there is a satisfactory pair: 
s = 2009, t = 2009^{2009}. 

We have
s(^{s2010}) = s^{s·(s2009)} = (s^{s})^{(s2009)}.
To get this in the form t^{t} we can let s = 2009. Similarly,
for any n>2, (s^{s})^{n} = t^{t} when s = n1 and t = (n1)^{(n1)}. 
So, for example, when n = 3, s = 2, and t = 2^{2} = 4 we again have 2^{(23)} = 4^{4}.
Further observations.
Kais and Lim investigated whether it might be possible to determine all solutions in the general settings of parts (b) and (c). For part (b) we fix a positive integer n, and
ask which pairs of integers s, t ≥ 2 satisfy (s^{s})^{n} = t^{t}. We shall obtain a partial solution to the question by adopting the very simple argument of Jakobsson, who was concerned only about finding a suitable approach to the problem as we stated it. It is clear that s and t must both have the same prime divisors; we will find all those solutions where there exists a common divisor a such that s = a^{k} and t = a^{m} for integers k and m. Because
(s^{s})^{n} = t^{t} holds if and only if a^{knak} = a^{mam},
it holds if and only if kna^{k} = ma^{m}. We deduce that there exists a solution pair s and t that are powers of the same integer a if and only if there exist integers k and m with m > k and
.
It is clear that if we take m to be any divisor greater than 1 of our given exponent n, there will be a solution with k = m – 1, namely
In particular, when we take m = n we get the solution s = (n – 1)^{n–1}, t = (n – 1)^{n} that we verified above. When n has proper divisors, there are further choices for m. Using n = 2010 as an example, we see that 2010 = 2·3·5·67, so n has 15 divisors m greater than 1, all of which lead to a solution:
m 
2010 
1005 
670 
402 
335 
201 
134 
67 
n/m 
1 
2 
3 
5 
6 
10 
15 
30 
s 
2009^{2009} 
2008^{1004} 
2007^{669} 
2005^{401} 
2004^{334} 
2000^{200} 
1995^{133} 
1980^{66} 
t 
2009^{2010} 
2008^{1005} 
2007^{670} 
2005^{402} 
2004^{335} 
2000^{201} 
1995^{134} 
1980^{67} 
m 
30 
15 
10 
6 
5 
3 
2 
n/m 
67 
134 
201 
335 
402 
670 
1005 
s 
1943^{29} 
1876^{14} 
1809^{9} 
1675^{5} 
1608^{4} 
1340^{2} 
1005 
t 
1943^{30} 
1876^{15} 
1809^{10} 
1675^{6} 
1608^{5} 
1340^{3} 
1005^{2} 
So, for example,
(1943^{29})^{(194329)}^{2010} = (1943^{29})^{30·67·(194329)} = (1943^{30})^{29·67·(194329)}
= (1943^{30})^{1943·(194329)} = (1943^{30})^{(194330)}
Kais used his computer to show that when n = 2010 there are just these 15 solutions, and no others. Both he and Lim showed that, on the other hand, there can be other possibilities for appropriate values of n. For example, when n = 11 we can take m = 11 and let k be 8 or 9 (as well as our canonical k = m – 1 = 10) :
Given n = 11, take m = 11 and k = 8; then kn/m = 8 = a^{11–8} = a^{3}; thus a = 2, s = 2^{8}, t = 2^{11}.
Given n = 11, take m = 11 and k = 9; then kn/m = 9 = a^{11–9} = a^{2}; thus a = 3, s = 3^{9}, t = 3^{11}.
Given n = 11, take m = 11 and k = 10; then kn/m = 10 = a; thus s = 10^{10}, t = 10^{11}.
Kais's computer showed that there were no further solutions when n = 11.
For further solutions to (s^{s})^{n} = t^{t}, we again set s = a^{k} and t = a^{m}. This now leads to
For n > 2 the choice of k = 1, m = n – 1, leads to the solution a = n – 1; whence, s = n – 1 and t = (n – 1)^{n–1}. Kais proved that there can be no solution when n = 2, and he showed (by computer) that the choice k = 1 produces the unique solution when n = 2010 that we have given above. Also here, for the exponent n = 11 there are three solutions with k = 1, and none for other values of k: for m = 8 we have s = 2, t = 2^{8}; for m = 9 we have s = 2, t = 2^{9}; and for m = 10 we have s = 2, t = 2^{10}.
Finally, we note that for fixed n, any solution to (s)^{sn }= t^{t} leads to a solution to (S^{S})^{n} = T^{T} by setting S = t and T = s^{n}: (S^{S})^{n} = ( t^{t})^{n} = (s^{(sn)})^{n} = s^{n(sn)} = (s^{n})^{(sn)} = T^{T}. You cannot go in the other direction however: we have seen that there are generally more solutions in part (b) than there are in part (c), so it is clear that solutions to (S^{S})^{n} = T^{T} do not lead in an obvious way to solutions of (s)^{sn }= t^{t}.
