pa2b2 small oddity

[Mark Hardman]

(02-01-2014 12:23 AM)Helge Gabert Wrote:  but why not include p=2 for the pa2b2 command? That is all I am asking.

I can see your point that, based on the help text, p=2 could be allowed.

Considering that the function explicitly states that the prime must meet the condition: \(p\equiv 1 \left ( mod\; 4 \right )\).

I am left to believe that it is a function for determining the two integers that solve Fermat's theorem on sums of two squares:

\(p=a^{2}+b^{2}\)

In addition to the requirement that \(p\equiv 1 \left ( mod\; 4 \right )\), the theorem also stipulates that p must be an odd prime number.

Mark Hardman