Vol. 10 No 6 - Pi Mu Epsilon

508 PI MU EPSILON JOURNAL
Since
sequence of ordered pairs (x;. Y;) where (x 1
, y 1 ) = (1, 1), (x 2 , y 2 ) = (3, 2)~ ,
x 5 x 1
and fori > 1, (x;.Y;) = (x;_ 1 + 2,
+--
s 7 + ... X;_ 1 + Yi-2).
Also solved by Miguel Amengual Covas, Charles Ashbacher, Frank P.
Battles, Paul S. Bruckman, James Campbell, Sandra Rena Chandler,
PROBLEMS AND SOLUTIONS 509
It then . follows that N r,p,a is (2r)! times the coefficient of x--p 1"n the
then we have
e~pans10n of Fp,a(x) as given by (11). The first few terms of this expansion
(ID terms of Sr,p,a) are
~ (-l)n-1 6 _ ~ (-l)n-1(11J3")2n-1 _ _ 1 1
L .......,.,~..,..,....,;,,----- - 6 L - 6tan _ = 1r.
n"'1 3(2n- 1 ) 12 (2n - 1) n"'1 2n- 1 {3
So,o,a = 1, Sl,O,a = a, S1,1,a = 1;
s2
,
0
'
a = 2a2 - a, s2
•
1
,
a = 3a - 2, s2
,
2
,a
= 1· '
Also solved by Frank P. Battles, Paul S. Bruckman, James Campbell,
Charles R. Diminnie, Russell Euler, George P. Evanovich, Jayanthi
Ganapathy, Robert C. Gebhardt, Richard I. Hess, Joe Howard, Thomas C.
s3,0,a = 6a 3 -9~ + 4a; s3 1 a= 12a 2 Leong, Peter A. Lindstrom, David E. Manes, Bob Prielipp, H.-J. Seiffert,
-22a + 11; s3 2 = 5a-4· s3 3
= 1. , ' ' ,a ' , ,a Rex H. Wu, and the Proposer.
Then the required probability is given by
886. [Spring 1996] Proposed by R. S. Luthar, University of Wisconsin
Center, Janesville, Wisconsin.
p s N,,IJ = ( 2r )M-2r S
Find the general solution in integers to the equation: - 8y + 7 = 0.
r .PIJ T - r ,p,p•
r,p r p
In terms of M = 2a the first few results are
I. Solution by David Tascione, student, St. Bonaventure University, St.
Bonaventure, New York.
P1,o,a = liM, P1,1,a = 11M 2 Suppose that (x, y) is a solution of: = 8y - 7. Since the right side is
; P2,o,a = 3(M- 1)/M3,
odd for any integer y, then x must be odd. We therefore take x = 2n + 1
P2,1,a = (6M- 8)/M 4 , P2,2,a = 11M 4 ; P3,0,a = 5(3M 2 - 9M + 8)/M 5 ,
P3,1,a = 15(3M 2 - llM + ll)/M 6 , P 3 2
= 3(5M- 8)/M6 p =
l!M6. • ,a • 3,3,a
for n an integer. Then
y = [(2n + 1 ) 2 + 7]/8 = 1 + n(n + 1)/2
which is an integer. It follows that (x, y) = (2n + 1, 1 + n(n + 1 )/2) is the
As mentioned, Part (a) is the special case of Part (b) for which p = 0.
Part (a) also solved by the Proposer.
general solution.
II. Solution by Skidmore College Problem Group, Saratoga Springs,
New York.
885. [Spring 1996] Proposed by Arthur Marshall, Madison, Wisconsin.
It is clear that y = (: + 7)/8, so if y is an integer, then
Evaluate the sum
E (-1)n-1 6
n"'1 3(2n- 1 ) 12 (2n- 1).
: = 1 (mod 8).
This congruence is true for all x where x is an odd positive integer. It can
Solution by Kenneth P. Davenport, Pittsburgh, Pennsylvania.
be shown by induction that all integer solutions may be written as a