Sum and product
Number Theory
Level
4
How many quadruplets of positive integers such that:
The answer is 9.
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From 2 equations, we have: a b − a − b = c + d − c d
or a b − a − b + c d − c − d = 0
or ( a − 1 ) ( b − 1 ) + ( c − 1 ) ( d − 1 ) = 2
Since a , b , c , d are positive integers, we only need consider 3 following cases:
Case 1: ( a − 1 ) ( b − 1 ) = 0 and ( c − 1 ) ( d − 1 ) = 2 . We get ( c , d ) = ( 2 , 3 ) or ( c , d ) = ( 3 , 2 ) . This implies to c + d = 5 , so ( a , b ) = ( 1 , 5 ) or ( a , b ) = ( 5 , 1 ) . Hence there are 4 satisfied quadruples in this case.
Case 2: ( a − 1 ) ( b − 1 ) = 1 and ( c − 1 ) ( d − 1 ) = 1 . So, a = b = c = d = 2 .
Case 3: ( a − 1 ) ( b − 1 ) = 2 and ( c − 1 ) ( d − 1 ) = 0 . This case is similar to case 1, and there are 4 satisfied quadruples in this case.
So, there are totally 9 quadruples satisfied.