Determine all possible pair(s) (M, N) of positive integers such that:

N^[√N] = M^M-1

Prove that these are the only possible pair(s) that exist.

Note: [x] denotes the greatest integer ≤ x.

10 for t=1 to 1000000
20      for n=1 to t-1
30        m=t-n
40        lhs=n^int(sqrt(n)):rhs=m^(m-1)
50        if lhs=rhs then ?n,m,lhs
60      next n
70 next t

 n       m     lhs (=rhs)
 1       1       1
 2       2       2
 8       4       64
Overflow in 40
?t
 888
OK

shows that for n+m < 888 there are only the three solutions:

(1,1)
(2,2)
(4,8)

A more sensibly written program (one that allows higher values) verifies no more solutions up to n = 48888, the point at which it was stopped:

10 for n=1 to 999999
20    v=n^int(sqrt(n))
30    v2=0:m=0
40    while v2<v
50       m=m+1:v2=m^(m-1)
60    wend
70    if v2=v then ?m,n,v
80 next
run

 m       n      value
 1       1       1
 2       2       2
 4       8       64
Break in 50
?n
 34644
OK
cont
Break in 50
?n
 48888
OK

 

Posted by Charlie on 2011-01-12 13:09:55