Let , where and are positive integers.
How many pairs of positive integers with for which every root of is an integer?
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In terms of a and b , f ( f ( x ) ) − x = x 4 − 2 a x 3 + ( a 2 − a + 2 b ) x 2 + ( 2 a 2 − 2 a b − 1 ) x + ( b 2 − a b + b ) . At first, this seems like a mess to factor, but suppose we had a root r of f ( x ) − x = 0 . Then we would also have a root of f ( f ( x ) ) − x = 0 since f ( r ) = r ⟹ f ( f ( r ) ) − r = f ( r ) − r = 0 . In other words, f ( x ) − x is a factor of f ( f ( x ) ) − x , so for some u and v , f ( f ( x ) ) − x = ( f ( x ) − x ) ( x 2 + u x + v ) = ( x 2 − ( a + 1 ) x + b ) ( x 2 + u x + v ) . Consider the constant term and the quadratic coefficient in the expansion of f ( f ( x ) ) − x above. Comparing those values to the factored form of f ( f ( x ) ) − x , we see that we must have b v = b 2 − a b + b , b + v − ( a + 1 ) u = a 2 − a + 2 b , which we can solve to find v = b − a + 1 , u = 1 − a . Now we are halfway to factoring f ( f ( x ) ) − x : f ( f ( x ) ) = ( x 2 − ( a + 1 ) x + b ) ( x 2 + ( 1 − a ) x + ( b − a + 1 ) ) . The roots of the two quadratic factors are a + 1 ± a 2 + 2 a − 4 b + 1 , a − 1 ± a 2 + 2 a − 4 b − 3 . The roots of f ( f ( x ) ) − x = 0 are integers if and only if the radicands are perfect squares. But the radicands differ by exactly 4, so this is only possible if a 2 + 2 a − 4 b + 1 = 4 , a 2 + 2 a − 4 b − 3 = 0 . Solving either equation for positive a gives the relation we want. The roots of f ( f ( x ) ) − x are integers exactly when a = − 1 + 2 b + 1 , b = n 2 − 1 ≥ 3 , where n is an integer. Based on this relation, there are 4 3 pairs ( a , b ) in the desired range.