Number Theory 719811.It is easy to guess that(x,y,z,t)=( 10 , 10 ,− 1 , 0 )is a solution. Because quadratic
Diophantine equations are usually simpler than cubic equations, we try to reduce the
given equation to a quadratic. We do this byperturbingthe particular solution that we
already know.
We try to find numbersuandvsuch that( 10 +u, 10 −u,−^12 +v,−^12 −v)is a
solution. Of course,vhas to be a half-integer, so it is better to replace it byw 2 , wherew
is an odd integer. The equation becomes
( 2000 +u^2 )−1 + 3 w^2
4= 1999 ,
which is the same as
w^2 − 80 u^2 = 1.This is a Pell equation. The smallest solution is(w 1 ,u 1 )=( 9 , 1 ), and the other positive
solutions are generated by
wn+un√
80 =(w 1 +u 1√
80 )n.This gives rise to the recurrence
(wn+ 1 ,un+ 1 )=( 9 wn+ 80 un,wn+ 9 un), n≥ 1.It is now easy to prove by induction that all thewn’s are odd, and hence any solution
(wn,un)to Pell’s equation yields the solution
(xn,yn,zn,tn)=(
10 +un, 10 −un,−1
2
+
wn
2,−
1
2
−
wn
2)
to the original equation.
(Bulgarian Mathematical Olympiad, 1999)
812.Consider first the case thatnis even,n= 2 k,kan integer. We have
(√
m+√
m− 1 )^2 k=( 2 m− 1 + 2√
m(m− 1 ))k.The term on the right-hand side generates the solution to Pell’s equation
X^2 −m(m− 1 )Y^2 = 1.If for a certainn,(Xn,Yn)is the corresponding solution, then choosep=X^2 n. Since
p− 1 =X^2 n− 1 =m(m− 1 )Yn^2 , it follows that
(√
m+√
m− 1 )^2 k=( 2 m− 1 + 2√
m(m− 1 ))k=Xn+Yn√
m(m− 1 )