Determine-all-pairs-x-y-of-integers-satisfying-1-2-x-2-2x-1-y-2-

Question Number 97746 by bemath last updated on 09/Jun/20

Determine all pairs (x, y)

of integers satisfying 1 + 2^{x} + 2^{2 x + 1} = y^{2}

Commented by john santu last updated on 09/Jun/20

2^{x} (1 + 2^{x + 1}) = (y - 1) (y + 1)

s h o w t h a t t h e f a c t o r s y - 1 a n d y + 1

a r e e v e n, exactly one of them

divisible by 4 . Hence x ⩾ 3 and one

of these factors is divisible by 2^{x - 1}

but not by 2^{x} . so y = 2^{x - 1} m + ϵ, m odd, ϵ = \pm 1

plugging this into the original

equation we obtain 2^{x} (1 + 2^{x + 1}) = {(2^{x - 1} m + ϵ)}^{2} - 1

= 2^{2 x - 2} m^{2} + 2^{x} m ϵ

or equivalently 1 + 2^{x + 1} = 2^{x - 2} m^{2} + m ϵ

\Leftrightarrow thus we have the complete

list solutions (x, y) : (0, 2), (0, - 2),

(4, 23), (4, - 23) .

Answered by Rasheed.Sindhi last updated on 18/Jun/20

1 + 2^{x} + 2^{2 x + 1} = y^{2}

2 {(2^{x})}^{2} + 2^{x} + 1 - y^{2} = 0

2^{x} = \frac{- 1 \pm \sqrt{1 - 8 (1 - y^{2})}}{4}

= \frac{- 1 \pm \sqrt{- 7 + 8 y^{2})}}{4}

8 y^{2} - 7 = u^{2}

y^{2} = \frac{u^{2} + 7}{8}

u^{2} = 0, 1, 4, 9, 16, \dots .

u^{2} = 1 \Rightarrow y = \pm 1

u^{2} = 25 \Rightarrow y = \pm 2

u^{2} = 121 \Rightarrow y = \pm 4

C o n t i n u e