The largest molecule is naturally ba^2, so we should consider killing it first.
By AB 2+B+7 ∣ AB 2+B+7
Get ab 2+b+7 ∣ a (ab 2+b+7)-b (ba 2+a+b) = 7a-b 2.
It is discussed in three cases: (1) 7a-b 2 = 0.
At this time, b = 7k, a = 7k 2, and k is an arbitrary positive integer, which meets the test conditions.
(2)7a-b^2<； 0
At this time, AB 2+B+7 ≤ B 2-7a of property 5 is divisible.
But ab 2+b+7 > b 2 >: B 2-7a, contradiction
(3)7a-b^2>； 0
Ab 2+b+7 ≤ 7a-b 2 of property 5 is divisible at this time.
But ab 2+b+7 > ab 2, 7a-b 2 < 7a, so b 2