ordered pairs (x,y)

\hspace{-16}$If $\bf{(x-8).(x-10)=2^y}$ where $\bf{x,y\in \mathbb{Z}}$. Then no. of ordered pairs of $\bf{\left(x,y\right)}$

1 Answers

2305

Shaswata Roy ·2013-02-28 20:23:06

For 8<x<10 , 2^y<0
Hence no solution for y can exist for 8<x<10

For x>10,
Both x-10 as well as x-8 must be a power of 2.
x-10 = 2^m & x-8 = 2ⁿ

or,2^m+10 = 2ⁿ+8
or,2ⁿ-2^m = 2
This is only possible when n=2 and m=1.

x = 12 , y = 3.

For x<8.

10-x = 2^m & 8-x = 2ⁿ

or,10 - 2^m= 8 - 2ⁿ
or,2^m-2ⁿ = 2
This is only possible when n=1 and m=2

x = 6 , y = 3.

Number of pairs = 2. (12,3) and (6,3)

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