Numbers with anticlosed sets of non-squares

[franktaw at netscape.net]

There is a well known result that, modulo any odd prime, every residue is the sum of two squares.  (The proof is fairly simple, using the pigeonhole principle.)  For p > 5, we can show that every residue is the sum of two non-zero squares.  If a + b = 1 with a and b non-zero squares, and r is any non-square, then ra + rb = r is a non-square as the sum of two non-squares.
 
For any odd p^n, n>1, p + p = 2p is a counterexample.  For 2^n, n>1, 3 + 3 = 6 is a counterexample.
 
It is straightforward to show that with n, m relatively prime, n*m can only have the property if n and m both do.
 
To complete the proof, we need only eliminate 15, where 2 + 3 = 5 is a counterexample.
 
Franklin T. Adams-Watters
16 W. Michigan Ave.
Palatine, IL 60067
847-776-7645
 
 
-----Original Message-----
From: jens at voss-ahrensburg.de


Let R be a ring. We will call a subset T of R *anticlosed* if for any
elements t1 and t2 of T, neither the sum t1 + t2 nor the product t1 * t2
lies in T.

For an integer n >= 2, let R := Z/nZ, and let T be the set of non-squares
of R. For which n is this set T anticlosed?

... Any idea on how to prove that no numbers beyond 10 have an
anticlosed set of non-squares?
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