Advanced book on Mathematics Olympiad

698 Number Theory

Fori=1 the property is obviously true, sincep 1 =2 and the consecutive terms of the
progression are odd numbers. Assume the property is true forp 1 ,p 2 ,...,pi− 1 and let
us prove it forpi.
Leta, a+d,a+ 2 d,...,a+(n− 1 )dbe the arithmetic progression consisting of
prime numbers. Using the inequalityd≥p 1 p 2 ···pi− 1 >pi, we see that if a term of
the progression is equal topi, then this is exactly the first term (in the special case of
p 2 =3, for which the inequality does not hold, the claim is also true because 3 is the
first odd prime). But ifa=pi, thena+pid, which is a term of the progression, is
divisible bypi, and the problem states that this number is prime. This means thata =pi,
and consequently the residues of the numbersa, a+d,...,a+(pi− 1 )dmodulopi
range over{ 1 , 2 ,...,pi− 1 }. By the pigeonhole principle, two of these residues must
be equal, i.e.,
a+sd≡a+td (modpi),

for some 0≤i<j≤pi−1. Consequently,a+sd−a−td=(s−t)dis divisible
bypi, and since|s−t|<pi, it follows thatdis divisible bypi. This completes the
induction, and with it the solution to the problem.
(G. Cantor)

757.We reduce everything modulo 3; thus we work in the ring of polynomials withZ 3
coefficients. The coefficients of bothP(x)andQ(x)are congruent to 1, so the reduced
polynomials areP(x)̂ =x

m+ (^1) − 1
x− 1 and
Q(x)̂ =xn+^1 −^1
x− 1. The polynomial
P(x)̂ still divides
Q(x)̂ ; thereforexm+^1 −1 dividesxn+^1 −1.
Letgbe the greatest common divisor ofm+1 andn+1. Then there exist positive
integersaandbsuch thata(m+ 1 )−b(n+ 1 )=g. The polynomialxm+^1 −1 divides
xa(m+^1 )−1, while the polynomialxn+^1 −1 dividesxb(n+^1 )−1 and so doesxm+^1 −1. It
follows thatxm+^1 −1 divides
xa(m+^1 )− 1 −(xb(n+^1 )− 1 )=xb(n+^1 )(xa(m+^1 )−b(n+^1 )− 1 )=xb(n+^1 )(xg− 1 ).
Hencexm+^1 −1 dividesxg−1. Becausegdividesm+1, this can happen only if
g=m+1. Therefore,m+1 is a divisor ofn+1, and we are done.
(Romanian Mathematical Olympiad, 2002)
758.We use complex coordinates, and for this, let
=cos
2 π
n
+isin
2 π
n

.

The vertices of the equiangular polygon should have coordinates

∑k

i= 0

σ(i)i,k= 1 , 2 ,...,n− 1 ,