10=m^p-n^q?

[Richard Guy]

Now in B16, I find that Golomb conjectured
that 6 is not the difference of two powererful
numbers (probably not the first to do so)
and Narkiewicz replied with 6 = 5^4*7^3- 463^2
aand noted that there were infinitely many such
representations.  Presumably it's still open
whether there are two pure powers that will do
the trick.     R.

On Mon, 7 Oct 2002, Richard Guy wrote:

> There's an enormous literature.  See D9 in UPINT
> for a sample.  There's a connexion with the
> ABC-conjecture -- B19.  I'd like to know if you
> fill in any of the gaps below, and would also
> like to see an extended list of gaps.  This
> poses a curious question: what about the sequence
> 
>  6, 14, 34, 42, 50, 58, 62, 66, 70, 78, ...
> 
> whose members are liable to evaporate down the
> years!  There are other examples, e.g.,
> numbers not known to be the sum of 3 cubes.
> 
> R.
> 
> On Mon, 7 Oct 2002, Don Reble wrote:
> 
> > > i've just speculated that somel integers can't be represented as 
> > > "difference of two full powers >=2"
> > 
> > One can immediately represent the odd numbers and the multiples of four,
> > as differences of two squares. Of the rest,
> > 
> > 	2 = 3^3 - 5^2
> >         6
> >        10 = 13^3 - 3^7 (Thanks, Neil)
> >        14
> >        18 = 3^3 - 3^2
> >        22 = 7^2 - 3^3
> >        26 = 3^3 - 1^2
> >        30 = 83^2 - 19^3
> >        34
> >        38 = 37^2 - 11^3
> >        42
> >        46 = 17^2 - 3^5
> >        50
> >        54 = 3^4 - 3^3
> >        ...
> > 
> > I haven't yet found reps for 6, 14, 34, 42, 50, 58, 62, 66, 70, 78, ...
> > Zak, do you have one for 6?
> > 
> > --
> > Don Reble       djr at nk.ca