(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**