818. Proposedby Angelo S. DiDornenico,Milfurd, MA.
Let \G"l denotea sequencedefinedby the reculrencerelation
Gn + 2:
Gn41 *
Gn,
where G6 and G1arc arbitrary real numbers.
(a) Show that for five consecutive elements in such a sequence,the difference of the
fourth power of the third element and the product of the other four elernents is a
constant;that is,
GX*, - G rGnalGn+zGn+q- K,
wh ereK isaconstant.
(b) Find a forrnula for K in terms of Ge and G1.
(c) Characterazethe sequencesfor which the constant K vanishes.
Proposed by Michael Andreoli, Miami-Dade College, Miami, FL.
Forn > l,evaluate
8L9.
rkun +t) ( k + 2) . - . ( k + n) '
820. Proposedby Greg Om.an,Ohio StateUniversity,Columbus,OH.
It is known that if two elementsc and b in anintegraldomain D aresuchthat a I b
that is, there existsa unit u e D btrchthat
andb la, then a andb arc associates;
a : ub. Show that the condition "D is an integral'domain"is necessary;that is, find
a commutativering with identity in which thereexist elementsx andy suchthat.r I y
andy | "r andx and y arenot associates.
SOLUTIONS
Roots of recursivelydefinedpolynomials
787. (Corrected) Proposedby Michel Bataille, Rouen,France.
Let {P"l be the sequenceof polynomialsdefinedby Ps(x) : 1, Pt(x) : x * 1, and
P"+z(x)-: xPn*t(x) - P"(x)
for all non-negativeintegersn. Find the roots of Pn.
Solution by Dionne Bailey, Elsie Campbell, Charles Diminnie, and Karl Havlak
(jointly), AngeloStateUniversiry,SanAngelo,TX.
This solutionfollows the discussionof the zerosof the ChebyshevPolynomialswhich
is givenin Theorem8.9 on pp.498499 of [1].
To begin,note that it is easily shownthat each Po is a polynomial of degreen. Also,
we will usethe following well-known trigonometricidentities:
sin34 : sinA(2 cos2A + l),
sinA* sinB:z r in( lJ t ) r o, r + g)
\2/
\2)
(1)
.
Q)