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From: ins_adsf@jhunix.UUCP (David S Fry)
Newsgroups: net.math
Subject: Re: A real problem! (Not a polar bear problem)
Message-ID: <1239@jhunix.UUCP>
Date: Wed, 20-Nov-85 12:16:21 EST
Article-I.D.: jhunix.1239
Posted: Wed Nov 20 12:16:21 1985
Date-Received: Sat, 23-Nov-85 04:57:42 EST
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Organization: Johns Hopkins Univ. Computing Ctr.
Lines: 27

> This yields the sequence of solutions: s = (2, 4, 14, 52, 214, ...)
>     				         h = (0, 2, 8,  30, 112, ...)
> 

This is incorrect.  The solutions for s are 
                 s = (2, 4, 14, 52, 194, 724, 2702, 10084,...)
Since one finds easily that n = s - 1, then n is
                 n = (1, 3, 13, 51, 193, 723, 2701, 10083,...).

Also, Mr. Silverman has only shown that these are necessary conditions,
not sufficient.  In other words, these numbers make 
                           sqrt(3 (n+3) (n-1))
an integer, but we do not necessarily know that they make
                      1/4 (n+1) sqrt(3 (n+3) (n-1))
an integer.

To see this, note that 2 always divides s (since s = 2 and s  = 4s   -  s  )
                                                  0         i     i      i-2
so 4 divides s^2.  Also, s^2 - 3h^2 = 4, or 3h^2 = s^2 - 4 so 4 divides 3h^2,
and hence 2 divides
                          sqrt(3 (n+3) (n-1)) = 3h.
And, n is always odd (n = s -1) so n+1 is even and divisible by 2.  Hence the
area is an integral.

NOW the solution is complete.

                                 David Fry