Given positive integer n, consider the set of numbers {n²+1, n²+2, ... (n+1)²}. If we pick two numbers x and y out of that set, how many different values can the product xy take?
There are 2n+1 numbers in the set, so there are 2n+1 choose 2 pairings
of different numbers plus 2n+1 perfect squares. This gives a total of
2n^2 + 3n +1 pairings, which is an upper bound to the number of
products we can generate. I can say this actually agrees with number of
unique products for the first few values of n.
As for a lower bound, we have 2n+1 different perfect squares ... I know this isn't particularly clever either :-)
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Posted by owl
on 2005-08-16 20:26:41 |
bounds
