2012:02: February Conundrum: Pairing Off

While daydreaming about what happens when two come together, it dawns on you that an interesting phenomenon occurs when two twoscome together!  Namely, whether you add them or multiply them, a pair of twos makes four.  Now you’re wondering if there are other pairs which fare so well when coupled.  You decide to take your matchmaking skills online and pose the following problem on your Facebook wall: are there other pairs of two positive integers such that the sum of the two numbers in the first pair equals the product of the two numbers in the second pair, and vice versa?  Now suppose that your privacy settings are such that all of your friends can see what every other friend has posted on your wall in this thread.  If you additionally promise a box of chocolates to every friend who supplies a different pair of pairs of positive integers than anyone who previously posted a solution, how many boxes of chocolates should you buy?

Congratulations to Dan Costanza, Daniel Phelps, and David Thompson for submitting correct solutions!

Solution [Due to Daniel Phelps]: So to start, I made the claim that a+b<ab for natural numbers a >2 and b>1 (basically excluding the cases where you have 1+a which is greater than 1*a and to avoid the case where you have 2+2=2*2). I used a proof by induction to prove this claim. Now while a+b<ab is true, a+b=cd and ab= c+d is a contradiction, because  ab = c+d < cd = a+b ==> ab<a+b.  This means that the only time that a group of numbers works out in the equation described is when one of the numbers is 1, or if they’re all 2.

So since were not looking for the solution with 2′s (you listed it in the problem), we can write:

1*b = c+d

b+1 = cd

a bit of substitution:


and some algebra:

c= (d+1)/(d-1)

Here I made a proof that says that c is not an integer for d>3 (to sum it up without all the formalism, d+1 is 2 more than d-1, and d-1 cant be a factor of d+1 if d-1 is more than half of d+1, so d-1 can’t be more than 2 and d can’t be more than 3.) Since d = 1 yields an undefined result, d is either 2 or 3, which makes c either 3 or 2.  Then b is c+d= 5 and we find that the only other cute couple of number pairs is 1,5 and 2,3.  Only one box of chocolates is necessary for this wall thread.