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<DIV>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.</DIV>
<DIV> </DIV>
<DIV>For any odd p^n, n>1, p + p = 2p is a counterexample. For 2^n, n>1, 3 + 3 = 6 is a counterexample.</DIV>
<DIV> </DIV>
<DIV>It is straightforward to show that with n, m relatively prime, n*m can only have the property if n and m both do.</DIV>
<DIV> </DIV>
<DIV>To complete the proof, we need only eliminate 15, where 2 + 3 = 5 is a counterexample.</DIV>
<DIV> </DIV>
<DIV>Franklin T. Adams-Watters<BR>16 W. Michigan Ave.<BR>Palatine, IL 60067<BR>847-776-7645</DIV>
<DIV> </DIV> <BR>-----Original Message-----<BR>From: jens@voss-ahrensburg.de<BR><BR>
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<DIV class=AOLPlainTextBody id=AOLMsgPart_0_b0e97355-ea5b-4097-8515-1b865c7895b1><PRE><TT>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?
</TT></PRE></DIV><!-- end of AOLMsgPart_0_b0e97355-ea5b-4097-8515-1b865c7895b1 --></DIV></DIV>
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