Vol. 8 No 7 - Pi Mu Epsilon

SIMONSON, Ean,! Tern State UnivmJULy wt Texuhfzuna, KENNETH M. WILKE,
Topeka, KS, and the. PROPOSER [becond boiu.tcon I. PLeLipp hound th^s,
phobien an, Exache 39 on page. 446 06 Clmjstal 's Textbook of Algebra,
voi. 1, 7th e.d; CheJLi,ea, 1964.
C& 0.
SotmLtLon by John V. Moohu, Cambtidge., Ma~bachu~e-fctA.
Clearly x > 0 implies y > 0 for each continued expression.
first expression is equivalent to
The
CoUe-ge., Uu-uig, TX, BOB PRIELIPP, Un&uUi^y
of^ W^c.onb.Ln-Obhkobh,
JOHN H. SCOTT, Macatutw. CoUege., S&wt Pa&, MN, KENNETH M. WILKE,
Topeka, KS, and the PROPOSER.
636. [Fall 19861 Phopohed by W&w. Btumbe~g, Cohd ~pfu.ng
FloLda.
a) Prove that if p is an odd prime, then 1 + p + p2 cannot be a
perfect square or a perfect cube.
*b) Is part (a) true when p is not prime?
I. So-tu-LLon 601 'the. ~quate. cae by Rob&
Hopatcong, New Jmey.
C. Ge-bhd,
2
Assume that 1 + p + p2 is a square, say k . Then
2
p2+p+(l-k)=0, SO p =-1+hk2-3.
2-2
The only square that is three less than another square is 1, which
occurs under the radical here when k = 1 or -1, in which case p = 0
or -1.
Thus p2 + p + 1 is never a square for prime p and is a square
for integral p only when p = -1 or 0.
11. So¥ttvLLo doh the. cube m e by KenneZh M.
kZflAaA.
2
Clearly p
for p = 0.
M-t^ke, Topeka,
+ p + 1 = s3 has integral solutions for p = -1 and
Let us assume the equation holds for any other integers p
and s. Then s > 1 and odd, and lpl > 8. Thus t = s - 1 is a
positive integer and, substituting t + 1 for s, we get that
(1
Since ]p
2
p(p + 1) = t(t + 3t + 3).
1 > s > t , then t divides p or p + 1. If t \ p, then we
write tj = p for some integer j. Then Equation (1) reduces to
The continued fraction is
y = l + % or y2 = y + x.
Y'
Hence both expressions represent the same parabola:
2
x = y -y, x, y > 0.
S-uluAiA bo.eu-t-t.on4 itfme. .~uamJWLe.d by JAMES E.
CAMPBELL,
UnLumLtq of, McMouAt., Columb-ta., RUSSELL EULER, Nowthwut ULiibouAt.
State. U&nu^ty, MmwUe., GEORGE P. EVANOVICH, So-wt PeAm
CoUige., Jmey Wy, NJ, ROBERT C. GEBHARDT, Hopcutaong, NJ, RICHARD
I. HESS, Rancho Pdob VeAdu, CA, PETER A. LINDSTROM, Novth Lake.
and applying the quadratic formula, we find that
2 2
We let F = (3 - j ) - 4(3 - j).
If, on the other hand, t \
p + 1, we take tk = p + 1, so that
p = tk - 1. Again substitute into Equation (1) and simplify to get .
2 2
tk - k = t +3t+3,
whose solution is