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### Pairs of Odd Integers

```
Date: 09/24/97 at 04:12:40
From: Brian Lam
Subject: Very Challenging Integer Problem

Dr. Maths,

I am a Year 9 student from Australia and I am asking you to help me
with a very hard and challenging integer problem.

Find all pairs of odd integers a and b that satisfy the following
equation : a + 128b = 3ab

Thank you!

Brian Lam
```

```
Date: 09/24/97 at 08:59:05
From: Doctor Rob
Subject: Re: Very Challenging Integer Problem

The trick for this kind of equation is to reduce it to the form
(A*a + B)*(C*b + D) = E, for some integers A, B, C, D, and E.  To that
end, bring all terms to the righthand side:

0 = 3*a*b - a - 128*b

Factor a from the first two terms:

0 = a*(3*b-1) - 128*b

This tells me that C = 3 and D = -1 in the above form.  Add 128/3 to
both sides to make the right side divisible by 3*b-1 exactly.

128/3 = a*(3*b-1) - 128/3*(3*b-1)
= (a-128/3)*(3*b-1)

Mulitply both sides by 3 to clear fractions:

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

This is in the desired form, and A = 3, B = -128, and E = 128.  Now
factor E = 128 into its prime power factors.

2^7 = (3*a-128)*(3*b-1)

Observe that both 3*b-1 and 3*a-128 must be factors of 2^7. Now use
the fact that a and b must both be odd integers to determine which
factors of 2^7 they can be. Don't forget that they could be negative!

-Doctor Rob,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 09/24/97 at 09:59:41
From: Doctor Anthony
Subject: Re: Very Challenging Integers Problem

Express b in terms of a:    a = b(3a-128)    so

a
b = --------
3a - 128

Now a and b are odd integers, and this equation shows us  a > 3a-128

128 > 2a    so a < 64

Also   3a-128 >= 1

3a >= 129    so a >= 43

Thus we have    43 <= a <= 63

If a = 43 then b = 43/(3x43 -128) = 43/1  = 43

So one pair of values is (43, 43)

We must stick to odd integers so the next value up for a is 45.

45         45
b = --------- =  -----   not an integer
3x45 -128      7

We can see that the numerator will increase by 2 each time and the
denominator by 6 (since it is 3a in the denominator), so the next few
calculations for b will have following results:

b = 47/13, 49/19,  51/25,  53/31, 55/37, 57/43, 59/49, 61/55, 63/61

and it is unnecessary to go further since the calculated values for b
will now become less than 1.

From the above calculation we see that there is just one pair of odd
integers satisfying the equation

a + 128 = 3ab.    and these are    a = 43,    b = 43.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra

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