# 2012:02: February Conundrum: Pairing Off

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:

c+d+1=cd

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.