720 Number Theory
=
√
p+√
p− 1 ,as desired.
This now suggests the path we should follow in the case thatnis odd. Write
(√
m+√
m− 1 )n=Un√
m+Vn√
m− 1.This time,(Un,Vn)is a solution to the generalized Pell equation
mU^2 −(m− 1 )V^2 = 1.In a similar manner we choosep=mUn^2 and obtain the desired identity.
(I. Tomescu,Problems in Combinatorics, Wiley, 1985)
813.First solution: This solution is based on an idea that we already encountered in the
section on factorizations and divisibility. Solving fory, we obtain
y=−x^2 + 4006 x+ 20032
3 x+ 4006.
To make the expression on the right easier to handle we multiply both sides by 9 and
write
9 y=− 3 x− 8012 −20032
3 x+ 4006.
If(x, y)is an integer solution to the given equation, then 3x+4006 divides 2003^2.
Because 2003 is a prime number, we have 3x+ 4006 ∈{± 1 ,± 2003 ,± 20032 }. Working
modulo 3 we see that of these six possibilities, only 1,−2003, and 2003^2 yield integer
solutions forx. We deduce that the equation from the statement has three solutions:
(− 1334 ,− 446224 ),(− 2003 , 0 ), and( 1336001 ,− 446224 ).
Second solution: Rewrite the equation as
( 3 x+ 4006 )( 3 x+ 9 y+ 8012 )=− 20032.This yields a linear system
3 x+ 4006 =d,3 x+ 9 y+ 8012 =−20032
d,
wheredis a divisor of− 20032. Since 2003 is prime, one has to check the casesd =
± 1 ,± 2003 ,± 20032 , which yield the above solutions.
(American Mathematical Monthly, proposed by Wu Wei Chao)