# Find all pairs of odd integer a and b wich satisfy the following equation: a+128b=3ab

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### 1 Answer

You should start subtracting a both sides such that:

`128b = 3ab-a`

You need to factor out a such that:

`128 b = a(3b-1)`

Since the number b is an odd integer, then 3b is also an odd integer and 3b-1 is an even integer.

Since 3b-1 is an even integer, then 3b-1 needs to be one of integer divisors of the even number 128 such that:

`D_128:{+-2;+-4;+-8;+-16;+-32;+-64;+-128}`

You may evaluate b solving the equations `3b - 1 = D_128` such that:

`3b-1 = 1 => b = 2/3 !in Z`

`3b-1 = -1 => b = 0 in Z`

`3b-1 =2 => b =1 in Z`

`3b-1 =-2 => b = -1/3 !in Z`

`3b-1 =4 => b = 5/3 !in Z`

`3b-1 = -4 => b = -1in Z`

`3b-1 =-8 => b = -7/3 !in Z`

`3b-1 = 8 => b =3 in Z`

`3b-1 = 16=> b =17/3 !in Z`

`3b-1 = -16 => b = -5 in Z`

`3b-1 =32 => b = 11 in Z`

`3b-1 = -32 => b =-31/3 !in Z`

`3b-1 = 64 => b = 65/3 !in Z`

`3b-1 =-64 => b = -21 in Z`

`3b-1 =128 => b =43 in Z`

`3b-1 = -128 => b =-127/3 !in Z`

**Hence, evaluating the odd integer values for a and b yields `{-21 ; -5 ; -1 ; 0 ; 1 ; 3 ; 11 ; 43} .` **