Quote:

> I have no doubt that a counter-example to the conjecture exists

>(see my comments in alt.math.recreational). I do object, however

>to the inclusion of zero as one of the valid integers in the

>conjecture. In fact it is very easy for example to find cubes that

>are sums of two squares (25 and 100 come to mind). My note

>in alt.math.recreational does NOT consider such cases.

...

Sorry if you found my original comment obscure.

If you don't like 0 as an integer, then the following (in J)

verifies that the sum of the fourth powers of 30, 120, 272, & 31

is 353^4:

+/30 120 272 315^4x

15527402881

353^4x

15527402881

The web page http://www.mathsoft.com/asolve/fermat/fermat.html

seems to suggest that this coincidence (mathematicians may prefer

the term "isolated result") can be found in Hardy & Wright's

"Theory of Numbers", which I don't possess.

Fourth powers modulo 16 are all either 0 or 1, showing that

one only need look for solutions of the form

(2a)^4 + (2b)^4 + (2c)^4 + (2d+1)^4 = (2e+1)^4,

but even with efficiencies like this a purely numerical search

for such formulae, particularly for higher powers, will need

a large workspace and a long time.

I couldn't view your original note since our newsreader claims

that "alt.math.recreational does not exist"---don't know why.

Here are a couple of other cute coincidences I found using J.

The first probably depends on deep properties of elliptic curves.

The second doesn't.

17j14 ": ,.(^. _104 + 12 28 396^4) % %: 10 18 58 NB. deep result?

3.14159590496940

3.14159265631841

3.14159265358979

666 A. 'moaned' NB. beastly but totally inconsequential result

daemon

Regards, Ewart Shaw.

--

Department of Statistics, University of Warwick, Coventry CV4 7AL, U.K.

http://www.warwick.ac.uk/statsdept/Staff/JEHS/