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Solution for March 2010

 The Problem: Here is a 3-part problem for month number 3: Can you find integers s and t, both greater than 1, for which (ss)2 = tt? Can you find integers s and t, both greater than 1, for which ( ss)2010 = tt? Can you find integers s and t, both greater than 1, for which s(s2010) = tt?

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) Mei-Hui 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 aNo, there are no integers s and t greater than 1 for which (ss)2 = tt.

Note first that (ss)2 = s2s. Should there exist such a pair s and t, then on the one hand because s,t >1 we get ss < s2s = tt, whence s < t; on the other hand, tt = s2s < (2s)(2s), so that t < 2s.  But t < 2s implies that st divides evenly into s2s = tt, 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 bYes, there are many possible pairs; for example,

 s = 1005 and t = 10052
and
 s = 20092009 and t = 20092010.

To discover the first pair we have

(ss)2010 = s2010s = s2·1005s = (s2)1005s,

which is in the form tt if we set s2 = 1005s; that is, s = 1005.

Comment. Note that this trick works for all even exponents except 2:

 for all integers k > 1, (ss)2k = tt when s = k and t = k2.

Try it with k = s = 2 and t = 4: (22)4 = 28 = 44.

Later we will see where the second pair comes from.  For now let us be satisfied in observing that

 for all integers n > 2, (ss)n = tt when s = (n – 1)n–1 and t = (n – 1)n.

Thus, in the smallest case n = 3 we have s = 22 = 4, t = 23 = 8, and

(ss)3 = (22)3·22 = (23)2·22 = (23)(23); that is, 412 = 88.

For our problem n = 2010, s = 20092009, and t = 20092010:

(ss)2010 = (20092009)2010·20092009 = (20092010)2009·20092009
= (20092010)(20092010) = tt.

In general,

(ss)n = ((n-1)n-1)n·(n-1)n-1 = ((n-1)n)(n-1)·(n-1)n-1 = ((n-1)n)((n-1)n) = tt.

Comment. We have here what might be called an anti-Fermat 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 cYes, there is a satisfactory pair:
 s = 2009, t = 20092009.

We have

s(s2010) = ss·(s2009) = (ss)(s2009).

To get this in the form tt we can let s = 2009.  Similarly,

 for any n>2, (ss)n = tt when s = n-1 and t = (n-1)(n-1).

So, for example, when n = 3, s = 2, and t = 22 = 4 we again have 2(23) = 44.

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 (ss)n = tt. 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 = ak and t = am for integers k and m.  Because (ss)n = tt holds if and only if aknak = amam, it holds if and only if knak = mam. 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 20092009 20081004 2007669 2005401 2004334 2000200 1995133 198066 t 20092010 20081005 2007670 2005402 2004335 2000201 1995134 198067

 m 30 15 10 6 5 3 2 n/m 67 134 201 335 402 670 1005 s 194329 187614 18099 16755 16084 13402 1005 t 194330 187615 180910 16756 16085 13403 10052

So, for example,

(194329)(194329)2010 = (194329)30·67·(194329) = (194330)29·67·(194329)
= (194330)1943·(194329) = (194330)(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 = a11–8 = a3; thus a = 2, s = 28, t = 211.
Given n = 11, take m = 11 and k = 9; then kn/m = 9 = a11–9 = a2; thus a = 3, s = 39, t = 311.
Given n = 11, take m = 11 and k = 10; then kn/m = 10 = a; thus s = 1010, t = 1011.

Kais's computer showed that there were no further solutions when n = 11.

For further solutions to (ss)n = tt, we again set s = ak and t = am. 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 = 28; for m = 9 we have s = 2, t = 29; and for m = 10 we have s = 2, t = 210.

Finally, we note that for fixed n, any solution to (s)sn = tt leads to a solution to (SS)n = TT by setting S = t and T = sn: (SS)n = ( tt)n = (s(sn))n = sn(sn) = (sn)(sn) = TT. 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 (SS)n = TT do not lead in an obvious way to solutions of (s)sn = tt.

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