## Re: [PrimeNumbers] a vey slow algorithm (enhanced)

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• In a message dated 30/11/2004 00:28:15 GMT Standard Time, sopadeajo2001@yahoo.es writes: If you want a proof of the theorem ,see Sloane A092541 Robin That s a
Message 1 of 2 , Nov 30, 2004
In a message dated 30/11/2004 00:28:15 GMT Standard Time,

If you want a proof of the theorem ,see Sloane A092541

Robin

That's a bit sneaky.
I thought you didn't have a proof and were asking for one... ah well.
If you had come up with that reference I wouldn't have claimed in my
yesterday's post to have finally proved your "little theorem".

However, that A092541 entry is a bit elliptical, to say the least -
incomplete, even.

For example, you don't mention the exceptions: 4,6,8.
(That they are the only ones is not easy to _prove_, as my post showed, I
think.)
You also don't state that a must be different from c and d.

It would also be interesting to know the background to your theorem, in
particular the following result, which is far from obvious, and which you include

>A general solution to m=a^2+b^2=c^2+d^2 for a known x=a+b+c+d is:
> c=(x(r-1)/2r)-a, d=(x+a(r-1))/(r+1) where r is a divisor of x/2.

(a) Is that taken from a published reference?
(b) it doesn't address the requirement that c (for example) should be >= 1 -
which, again, was something my proof needed to take careful account of.
And so on.

(BTW it's an odd coincidence that you and I should have quite independently
used the symbol "r" for very much the same thing - a factor of x/2 - when
there are all those other letters available! )