--- In

primenumbers@yahoogroups.com,

"marku606" <mark.underwood@...> wrote:

>> Just checked up to 300,000 for numbers of the form 8n+5:

6531 of such numbers are primes or prime powers, and they

can all be written as the sum of two relatively prime

squares only once. It appears that numbers of the form 8n+5

which are not prime or prime powers cannot be written as the

sum of two relatively prime powers only once.

For number of the form 8n+1: 6572 of such numbers are

primes or prime powers, but only 6510 of these can be

written as the sum of two relatively prime squares only

once. The sixty two that cannot are all primes powers,

specifically, primes raised to an even power.

As in the case for numbers of the form 8n+5, it appears

that numbers of the form 8n+1 which are not prime or prime

powers cannot be written as the sum of two relatively prime

powers only once. <<

All this is very simply explained

by unique factorization in K(sqrt(-1)).

1) A number N is the sum of two squares if and only if

every prime divisor of N that is congruent to 3 mod 4

occurs to an even power.

2) An odd number N is the sum of two coprime squares in precisely

one way if only if N is a prime congruent to 1 mod 4

or a power of such a prime.

Note that (1) is a consequence of Theorem 252 of Hardy and Wright,

that the primes of K(sqrt(-1)) are: 1 + I, the rational primes

congruent to 3 mod 4, the factors a + b*I of the rational primes

congruent to 1 mod 4, and the associates of these primes.

To understand (2), suppose that

p = norm(a+I*b) = a^2 + b^2 and

q = norm(c+I*d) = c^2 + d^2

are distinct rational primes congruent to 1 mod 4.

Then we may write

p*q = norm((a+I*b)*(c+I*d)) = (a*c-b*d)^2 + (b*c+a*d)^2

p*q = norm((a+I*b)*(c-I*d)) = (a*c+b*d)^2 + (b*c-a*d)^2

in two essentially different ways.

But for any power k of p, the only sum of coprime squares is

p^k = norm((a+I*b)^k)

since if we try something like

p^k = norm((a+I*b)^(k-j)*(a-I*b)^j)

with k > j > 0, then p will divide both of the resultant squares.

David