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Another case of divisibility (Posted on 2009-02-15) Difficulty: 2 of 5
Determine all pairs (a, b) of positive integers such that ab2 + b + 7 divides evenly a2b + a + b.

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Solution Alternative Methodology Comment 3 of 3 |
(In reply to Analytic Solution by K Sengupta)

A slightly different methodology is as follows:

As before, (a^2b+a+b)/(ab^2+b+7) = p(say).  Then

->  p*b- a = (b^2-7a)/(a*b^2+b+7), since each of s, a and b are integers, it follows that the expression to the right must be likewise. Also, the numerator is < b^2, but the denominator is > b^2, so that:  p*b- a < 1

Case (I): (p*b-a) < 1.  
In this case, we have:
7a-b^2>   a*b^2+b+7
-> 7a > (a+1)*b^2 > a*b^2
-> b^2 < 7, so that: b= 1, or 2
If b=1, then:
p*b-a = (1-7a)/(a+8) = 57/(a+8)  - 7


Since  p*b-a is an integer, it follows that (a+8) divides 57. Since the integers >= 8 that divides 57 are 19 and 57, we must have;
a+8 = 19, 57, giving: a = 11, 49

Whenever b=2, we have
4(p*b-a) = 79/(4a+9) 7. Clearly, the expression on the lhs is an integer, so that the corresponding expression on the rhs is likewise.

Now, the only integer divisor of 79 that is >= 9/4 is 79. But the equation:  4a+9 = 79 do not lead to any integer value for a. This is a contradiction.

Case II: (p*b a) = 0

Then, we have: p= a/b, so that:
(a^2b+a+b)/(ab^2+b+7) = a/b
-> b^2 = 7a (upon simplification)
-> b/7 = a/b
-> b/7 = p
-> b= 7*p, so that:
a = 7*p^2

So, (a, b) = (7*p^2, 7p) for any given integer p satisfy all the given conditions.

Consequently,

(a, b) = (11,1), (49,1), and (7*(p^2), 7p), whenever p is any given positive integer.

Edited on February 15, 2009, 11:55 pm
  Posted by K Sengupta on 2009-02-15 14:33:50

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