ON DIOPHANTINE EQUATION
X 2 = 2Y 4 − 1
Florentin Smarandache University of New Mexico 200 College Road Gallup, NM 87301, USA E-mail: smarand@unm.edu Abstract: In this note we present a method of solving this Diophantine equation, method which is different from Ljunggren’s, Mordell’s, and R.K.Guy’s. In his book of unsolved problems Guy shows that the equation x 2 = 2y 4 − 1 has, in the set of positive integers, the only solutions (1,1) and (239,13) ; (Ljunggren has proved it in a complicated way). But Mordell gave an easier proof. We’ll note t = y 2 . The general integer solution for x 2 − 2t 2 + 1 = 0 is ⎧ xn +1 = 3xn + 4t n ⎨ ⎩t n +1 = 2xn + 3t n for all n ∈N , where (x0 , y0 ) = (1, ε ) , with ε = ±1 (see [6]) or ⎛ xn ⎞ ⎛ 3 4 ⎞ ⎛ 1 ⎞ ⎜⎜ ⎟⎟ = ⎜ ⎟ ⋅ ⎜ ⎟ , for all n ∈ N , where a matrix to the power zero is 2 3 t ⎝ ⎠ ⎝ε ⎠ ⎝ n ⎠ equal to the unit matrix I . ⎛ 3 4⎞ Let’s consider A = ⎜ , and λ ∈ R . Then det(A − λ ⋅ I ) = 0 implies ⎝ 2 3⎟⎠ n
λ1,2 = 3 ± 2 , whence if v is a vector of dimension two, then: Av = λ1,2 ⋅ v . Let’s consider
2⎞ ⎛2 P=⎜ ⎟ ⎝ 2 - 2⎠
and
⎛3+ 2 2 0 ⎞ D=⎜ ⎟ . We have 3 -2 2 ⎠ ⎝ 0
P −1 ⋅ A ⋅ P = D , or ⎛ ⎜ An = P ⋅ D n ⋅ P −1 = ⎜ ⎜ ⎜ ⎝
(
where a = 3 + 2 2
)
1 ( a + b) 2 2 ( a − b) 4 n
(
⎞ 2 (a − b) ⎟ 2 ⎟, ⎟ 1 ( a + b) ⎟ 2 ⎠
)
n
and b = 3 − 2 2 .
Hence, we find:
1