Advanced book on Mathematics Olympiad

Number Theory 717

805.One can verify thatx= 2 m^2 +1 andy= 2 mis a solution.
(Diophantus)

806.We will search for numbersxandyfor which 2x^2 =a^2 and 2y^2 = 2 a, so that
1 + 2 x
2

• 2 y
  2
  =(a+ 1 )^2. Thenx= 2 zfor some positive integerz, and

a= 22 z

2

= 2 y

(^2) − 1
.
This leads to the Pell equation
y^2 − 2 z^2 = 1.
This equation has infinitely many solutions, given by
yn+zn

√

2 =( 3 + 2

√

2 )n,

and we are done.
(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette), proposed by
M. Burtea)

807.The Pell equationx^2 − 2 y^2 =1 has infinitely many solutions. Choosen=x^2 −1.
Thenn=y^2 +y^2 ,n+ 1 =x^2 + 02 , andn+ 2 =x^2 + 12 , and we are done.
(61st W.L. Putnam Mathematical Competition, 2000)

808.In other words, the problem asks us to show that the Diophantine equationx^2 − 2 = 7 y
has no positive solutions. A reduction modulo 8 makes the right-hand side equal to(− 1 )y,
while the left-hand side could only be equal to− 2 ,− 1 ,2. This means thatymust be
odd,y= 2 z+1, withzan integer.
Multiplying by 7y = 72 z+^1 and completing the square, we obtain the equivalent
equation

( 72 z+^1 + 1 )^2 − 7 ( 7 zx)^2 = 1.

Let us analyze the associated Pell equation

X^2 − 7 Y^2 = 1.

Its fundamental solution isX 1 =8,Y 1 =3, and its general solution is given by

Xk+Yk

√

7 =( 8 + 3

√

7 )k,k= 1 , 2 ,....

SubstitutingX= 72 z+^1 +1 andY= 7 zx, we obtain

72 z+^1 + 1 = 8 k+

(

k

2

)

8 k−^2 · 32 · 7 +

(

k

4

)

8 k−^4 · 34 · 72 +···,