Smallest integer N = a^(a+2*b) = b^(b+2*a) ?

 

Find the smallest integer N which satifies:

 

SMALLEST N

 
 
I did not specify the domains of N, so we could have a discussion and consider several cases
 

Case #1

N   is a positive integer,   and   a, b   integers.

The least value of N is 1.   This occurs for   (a, b) = (-2, 1)

 

Case #2    N,   a,   b   positive integers

Let   b = k*a,   with rational   k > 1,   then
SMALLEST N 2

It is clear that a and b are integers if and only if both k and   3/(k-l)   are integers,
that is. if and only if   k = 2   or   k = 4.

k = 2   —>   (a, b) = (16, 32)   —>   N \; = \; 16^{80} \; = \; 32^{64} \; = \; 2^{320}

and

k = 4   —>   (a, b) = (16, 64)   —>   N \; = \; 16^{144} \; = \; 64^{96} \; = \; 2^{576}

 

The smallest integer is    N \; = \; 2^{320}

 

Now consider case #3   N positive integer and   a, b   positive real numbers.

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

2 Responses to Smallest integer N = a^(a+2*b) = b^(b+2*a) ?

  1. paul says:

    If we allow negative values for a or and b the smallest value of N is 1, if not then it is quite large. Format is {a, b, N}

    {-2, 1, 1}

    {16,32,2135987035920910082395021706169552114602704522356652769947041607822219725780640550022962086936576}.

    I’m not 100% but I recon these are the only solutions for a and b when N is an integer.
    {-4,32}
    {-2,1}
    {16,32}
    {16,64}

    There are a few more where a^(a+2b) = b^(b+2a) but N is fractional.

    Paul.

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