718 Number Theory
7 z=(
k
1)
8 k−^1 · 3 +(
k
3)
8 k−^3 · 33 · 7 +(
k
5)
8 k−^5 · 35 · 72 +···.Let us compare the power of 7 ink=
(k
1)
with the power of 7 in( k
2 m+ 1)
7 m,m>1. Writ-ing
( k
2 m+ 1)
7 m=^7mk(k− 1 )···(k− 2 m− 1 )
1 · 2 ···k , we see that the power of 7 in the numerator grows
faster than it can be canceled by the denominator. Consequently, in the second equality
from above, the power of 7 in the first term is less than in the others. We thus obtain that
7 zdividesk. But then 8k> 87 z> 72 z+^1 , and the first inequality could not hold. This
shows that the equation has no solutions.
(I. Cucurezeanu)
809.Expanding the cube, we obtain the equivalent equation 3x^2 + 3 x+ 1 =y^2. After
multiplying by 4 and completing the square, we obtain( 2 y)^2 − 3 ( 2 x+ 1 )^2 =1, a Pell
equation, namely,u^2 − 3 v^2 =1 withueven andvodd. The solutions to this equation are
generated byun±vn
√
3 =( 2 ±
√
3 )n, and the parity restriction shows that we must select
every other solution. So the original equation has infinitely many solutions generated by
2 yn±( 2 xn+ 1 )√
3 =( 2 ±
√
3 )( 5 ± 4
√
3 )n,or, explicitly,
xn=( 2 +
√
3 )( 5 + 4
√
3 )n−( 2 −√
3 )( 5 − 4
√
3 )n− 1
2,
yn=( 2 +
√
3 )( 5 + 4
√
3 )n+( 2 −√
3 )( 5 − 4
√
3 )n
2.
810.One family of solutions is of course(n, n),n∈N. Let us see what other solutions
the equation might have. Denote bytthe greatest common divisor ofxandy, and let
u=xt,v=yt. The equation becomest^5 (u−v)^5 =t^3 (u^3 −v^3 ). Hence
t^2 (u−v)^4 =u^3 −v^3
u−v=u^2 +uv+v^2 =(u−v)^2 + 3 uv,or(u−v)^2 [t^2 (u−v)^2 − 1 ]= 3 uv. It follows that(u−v)^2 divides 3uv, and sinceu
andvare relatively prime andu>v, this can happen only ifu−v=1. We obtain the
equation 3v(v+ 1 )=t^2 −1, which is the same as
(v+ 1 )^3 −v^3 =t^2.This was solved in the previous problem. The solutions to the original equation are then
given byx=(v+ 1 )t,y=vt, for any solution(v, t)to this last equation.
(A. Rotkiewicz)