# Find all pairs of odd integers a and b which satisfy the folowing equation: a + 128b = 3ab

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### 1 Answer

I suggest to you to subtract `a` both sides such that:

`128b = 3ab - a`

Factoring out `a` to the right yields:

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

The problem provides the information that the number `b` is an odd integer, hence `3b` is also an odd integer and `3b -1` is an even integer.

Since `128 b = a(3b - 1)` and `3b -1` is an even integer, then `3b -1` must be one of integer divisors of the even number 128 such that:

`D_128:{+-1 ; +-2 ; +-4 ; +-8 ; +-16 ; +-32 ; +- 64 ; +-128}`

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

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

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

`3b - 1 = -2 =gt 3b = -2 + 1 =gt 3b = -1 =gt b = -1/3 !in Z `

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

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

`3b - 1 = 8 =gt 3b = 9 =gt b = 3 in Z `

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

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

`3b - 1 = -16 =gt 3b = -15 =gt b = -5in Z `

`3b - 1 = 32 =gt 3b = 33 =gt b = 11in Z`

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

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

`3b - 1 = -64 =gt 3b = -63 =gt b = -21 in Z`

`3b - 1 = 128 =gt b = 129/3 =gt b = 43 in Z `

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

Hence, b could take the following odd integer values `{-21,-5,-1,0,1,3,11,43}.` Notice that the relation provided by the problem `128 b = a(3b - 1)` states that `a` needs to divide or be equal to `b` , hence `a` could take the same values as `b` .

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

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