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Posting-Version: version B 2.10.2 9/18/84; site bonnie.UUCP
Path: utzoo!watmath!clyde!bonnie!rhm
From: rhm@bonnie.UUCP (Bob Morris)
Newsgroups: net.math
Subject: Re: A number theory problem
Message-ID: <549@bonnie.UUCP>
Date: Tue, 3-Sep-85 11:38:19 EDT
Article-I.D.: bonnie.549
Posted: Tue Sep  3 11:38:19 1985
Date-Received: Wed, 4-Sep-85 06:57:25 EDT
References: <388@aero.ARPA>, <946@oddjob.UUCP> <447@ttidcb.UUCP>
Organization: AT&T Bell Laboratories, Whippany NJ
Lines: 51

> 
> > Misspending part of a summer on this sort of question led me to
> > observe that the number 221*5^n seems to be expressible as a sum
> > of two squares 2n+2 different ways for n through at least 10 or
> > so.  Is it true for all n?  If so, why?
> 
> Yes, it is true for _all_ n.  In fact the more general problem can be
> phrased as follows:
> 
> 			    n
> 			_________
> 	    |\    |       |   |     |/
> 	    | \   |       |   |   _ |\i
>    Let      |  \  | =     |   |  |_)
> 	    |   \ |       |   |  |  i
> 	    |    \|       |   |
> 			   i=1
> 
> 
>    where each p  is a prime of the form 4x+1 and further
> 	       i
>    let f(N) be the number of _different_ divisors of N (1 is
>    considered a divisor) less than or equal to the square root of N.
> 
>    Then N can be represented as the sum of 2 squares in exactly f(N) ways.
> 
>    -----------------------------------------------------------------------
> 
>    First proof wins a box of cheerios.  My proof is over 200 lines long so
>    I'm not posting at this time --- margin is too small to contain it  :-)
>    Hint: It can be solved by elementary means, that is to say algebraically
> 	 rather than analytically!
> 
>    p.s. Don't worry about misspending a summer on this problem, I've
> 	wasted my whole life (48 and still counting) trying to prove
> 
> 			x^n + y^n = z^n
> 
> 	has no integral solutions for n>2   :-(
> 						   Enjoy!

*** REPLACE THIS LINE WITH YOUR MESSAGE ***
Spending a summer to come up with 221*5^n is overkill.
It works as stated and the proof is available in any non-trivial
book on number theory.
But 13*5^n works equally well, and, indeed, 5^n is not bad either.

On another subject that came up recently, there are numbers that
can be expressec in k ways as the sum of two cubes, for any value
of k.  The proof is (strangely enougn) constructed and can be
found somewhere toward the end of Hardy and Wright.