Binary neighbors perfect polynomials, instead of consecutive - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T18:14:48Z http://mathoverflow.net/feeds/question/52649 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52649/binary-neighbors-perfect-polynomials-instead-of-consecutive Binary neighbors perfect polynomials, instead of consecutive Luis H Gallardo 2011-01-20T16:41:52Z 2011-01-21T08:42:04Z <p>Inspired by Andrei's nice solution of 52609, (namely <code>consecutive</code> perfect polynomials): Denote by $A$ the full ring of polynomials in one variable $t$ over the finite field with $2$ elements. For any polynomial $P \in A$ define $$\sigma(P) = \sum_{d \mid A} d.$$ A polynomial $P \in A$ is called <code>perfect</code> if $$P = \sigma(P).$$ If $P$ has no roots in $GF(2)$ then $P$ is <code>odd</code>, otherwise it is <code>even</code>.</p> <p>Andrei proved: </p> <p>$P$ is odd perfect iff $P$ is perfect and $P$ is a square in $A,$ and deduced:</p> <p>$P$ and $P+1$ cannot both be perfect.</p> <p>Observe that one of $P,P+1$ is odd while the other is even.</p> <p>Take now a polynomial $P \in A.$ What is the <code>next</code> polynomial of the same <code>type</code>? i.e., having the same <code>parity</code>. Seems that the following definition is appropriate for this:</p> <p>Call <code>neighbors</code> two polynomials $P,Q \in A$ if $\deg(P)>2$ and ($Q=P+t(t+1)$ or $P=Q+t(t+1)$).</p> <p>Question: There are neighbors polynomials $P,Q \in A$ such that $P$ and $Q$ are both perfect ? </p>