Find 4 different positive integers A,B,C,D for which:
A+B = C*D and
A*B = C+D
How many sets of 4 numbers can you find? Prove that only those sets exist.

(In reply to

answer by K Sengupta)

The given equations are:

A+B = C*D and A*B = C+D

This yields:

AB + CD = A+B+C+D

Or, (A-1)(B-1) + (C-1)(D-1) = 2

Since A, B, C and D are positive integers, it follows that (A-1), (B-1), (C-1) and (D-1) are non negative integers.

Thus, we arrive at the following cases.

*Case (i)*:(A-1)(B-1) = (C-1)(D-1) = 1

This gives (A,B, C, D) = (2,2,2,2) which is a contradiction as all the four numbers are equal.

*Case (ii)*: (A-1)(B-1) =2; (C-1)(D-1) = 0

This gives (A, B) = (2,3); (3,2), and:

either C=1, or D = 1 or C=D=1

C=D=1 is untenable as C+D = 2 != 5 = AB.

C=1 gives D = 6-1 = 5 from the original eqiuation

Likewise, D=1, gives C=5

Thus, (A, B, C, D) = (2,3,1,5); (2,3,5,1); (3,2,1,5); (3,2,5,1)

*Case (iii)*: (A-1)(B-1) = 0; (C-1)(D-1) = 2

Following similar arguments as in case (ii), we obtain:

(A, B, C, D) = (1,5, 2, 3); 5,1, 2, 3)); (1,5, 3, 2); (5,1, 3, 2)

Consequently, (A,B,C,D)= (2,3,1,5); (2,3,5,1); (3,2,1,5); (3,2,5,1);(1,5, 2, 3); (5,1, 2, 3); (1,5, 3, 2); (5,1, 3, 2).

*Edited on ***May 20, 2007, 3:39 pm**