Among pairs of numbers whose sum is 16, 8*8=64 is the greatest possible product. However, if we allow for a sum of 17 then there are 3 distinct ways of achieving a higher product with two positive integers: 9*8, 10*7, and 11*6 are all greater than 64.

Among pairs of numbers whose sum is 2n, n*n=n^{2} is the greatest possible product. However, if we allow for a sum of 2n+1 then there are 2012 distinct ways of achieving a higher product with two positive integers.

Find the minimum value of n.

First we know that for a number of the form 2n+1, n being integer,and if the two parts of the product being integer, the maximum value occurs at n*(n+1). from this point we construct the inequality:

n*(n+1)*(n-a)/n*(n+a+1)/(n+1) - n^2 > 0

→ (n-a) (a+n+1)-n^2 > 0

→-a^2-a+n > 0

Solving for a at -a^2-a+n = 1

we get for positive solutions:

a= 1/2 (Sqrt[4 n-3]-1)

Now, if we take the floor of the solution we have the point at the series stating at (n-a)*(n+1+a) where this expression stops being greater that n^2.So the unique integer pairs that satisfy this are equal to Floor[a] + 1(we have to include a=0),solving this for 2012 we have: n= 4046133.

*Edited on ***June 13, 2012, 6:58 am**