an observation...

[David Wilson]

Your observation is true.  Given that triviality is a subjective attribute,
I took a poll which included my wife, son, daughter, my cat Kitty, and
John Conway as to whether this observation was trivial.  60% of those
surveyed agreed it was nontrivial, 40% did not respond.  Error is
within 101%.

The presumably nontrivial part is that if n >= 1 is the sum of two squares,
then every prime divisor p == 3 (mod 4) of n must have even exponent in
the prime factorization of n.  (Other primes may have any exponent).

Blum numbers have the prime factorization b = pq where p != q and
p == q == 3 (mod 4).  From this, b == 1 (mod 4) follow easily.  Also,
because p == 3 (mod 4) and occurs with odd exponent 1 in the prime
factorization of b, the paragraph above allows us to conclude that b
is not the sum of two squares.

Consequently, every Blum number is == 1 (mod 4) and is not the sum of
two squares, which places it in A084109, as observed.

----- Original Message ----- 
From: "Ralf Stephan" <ralf at ark.in-berlin.de>
To: <ham>; "SeqFan" <seqfan at ext.jussieu.fr>
Sent: Sunday, April 17, 2005 4:20 AM
Subject: an observation...


> The first sequence seems a subset of the second. Is this trivial?
>
> %S A016105 
> 21,33,57,69,77,93,129,133,141,161,177,201,209,213,217,237,249,253,301,
> %N A016105 Blum numbers: of form P*Q where P&Q are distinct primes 
> congruent to 3 (mod 4).
>
> %S A084109 
> 21,33,57,69,77,93,105,129,133,141,161,165,177,189,201,209,213,217,237,
> %N A084109 n is congruent to 1 (mod 4) and is not the sum of two squares.