Solve-for-integers-x-x-4-5-3-y-1-

Question Number 157902 by HongKing last updated on 29/Oct/21

Solve for integers :

x \cdot (x + 4) = 5 \cdot (3^{y} - 1)

Commented by Rasheed.Sindhi last updated on 01/Nov/21

3^{y} = \frac{x \cdot (x + 4)}{5} + 1 = \frac{x (x + 4) + 5}{5}

3^{y} = \frac{x (x + 4) + 5}{5}

x = 3 m :

3^{y} = \frac{3 m (3 m + 4) + 5}{5} = \frac{{9 m}^{2} + 12 m + 5}{5}

m = 0 : 3^{y} = 1 = 3^{0} \Rightarrow y = 0 \Rightarrow x = 3 m = 0

⋮

x = 3 m + 1 :

3^{y} = \frac{x (x + 4) + 5}{5} = \frac{(3 m + 1) (3 m + 5) + 5}{5}

= \frac{{9 m}^{2} + 18 m + 5}{5} \Rightarrow 5 ∣ m

= \frac{(3 m + 5) (3 m + 1)}{5}

⋮

x = 3 m + 2

⋮

Answered by Rasheed.Sindhi last updated on 29/Oct/21

x \cdot (x + 4) = 5 \cdot (3^{y} - 1) \dots \dots \dots A

\Rightarrow 5 ∣ x \lor 5 ∣ (x + 4)

5 ∣ x : L e t x = 5 m; m \in Z

A \Rightarrow 5 m (5 m + 4) = 5 (3^{y} - 1)

\Rightarrow m (5 m + 4) = 3^{y} - 1

\Rightarrow {5 m}^{2} + 4 m + 1 - 3^{y} = 0

m = \frac{- 4 \pm \sqrt{16 - 20 (1 - 3^{y})}}{10}

20 . 3^{y} - 4 is perfect square

m = \frac{- 4 \pm \sqrt{20 . 3^{y} - 4}}{10} \Rightarrow y = 0

m = 0, - \frac{4}{5}

x = 5 m = 0, - 4

x = 0, y = 0

x = - 4, y = 0

5 ∣ (x + 4) : Let x + 4 = 5 m

x = 5 m - 4

(5 m - 4) (5 m - 4 + 4) = 5 (3^{y} - 1)

{5 m}^{2} - 4 m = 3^{y} - 1

{5 m}^{2} - 4 m + 1 - 3^{y} = 0

m = \frac{4 \pm \sqrt{16 - 20 (1 - 3^{y})}}{10}

m = \frac{4 \pm \sqrt{20 . 3^{y} - 4}}{10} \Rightarrow y = 0

= \frac{4 \pm 4}{10} = \frac{4}{5}, 0

x + 4 = 5 m

m = 0 \Rightarrow x = - 4

m = \frac{4}{5} \Rightarrow x = 5 m - 4 = 0

(- 4, 0), (0, 0)

Commented by HongKing last updated on 29/Oct/21

alot thankyou sir cool