Suppose A=a^2+b+a; B=ab^2+b+7

Now if B divides A then it also divides bA-aB.

We are hence looking for those pairs (a,b) for which B divides bA-aB=b^2-7a

Since a1 we have  b^2 <=ab^2

Therefore b^2-7a<ab^2+b+7=B

Let b^2-7a>=0 first , then B is dividing a non--ve integer smaller than B itself and thus the later b^2-7a=0

then b^2=7a

Thus b is a multiple of 7 can be written as b=7c

Hence 4ac^2-7a=0 and a=7c^2

where c is a +ve integer.

These pairs (a,b)=(7c^2,7c) already satisfy the conditions since

A=49c^4.7c+7c^2+7c

  = 7c(49c^4+c+1)

B=7c^2.49c^2+7c+7

  =7(49c^4+c+1)

i.e. A=cB

Assume that b^-7a<0; its opposite the +ve integers 7a-b^2<7a is divisible by B. b now cannot exceed 2 otherwise B=ab^2+b+7>=9a+10

Thus B can not divide a +ve number less than 7a.

Hence b=1 or b=2

If b=1,then 7a-b^2=7a-1

If this is a multiple of B=a+8

then 7a-1=7(a+8)-57 shows that B has to divide 57.

thus a =11 or 49

For b=2 we get no soln.

The sols are the pairs (7c^2,7c) and the two extras :(11,1) and (49,1).

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