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

Find the smallest integer N which satifies:

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

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.

math grad - Interest: Number theory
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### 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.

• benvitalis says: