Hello druvid fae
Consider the case . Clearly there is no solution unless , since , and one solutions otherwise.
Now consider .
Since and , if , there is no solution; and there is no solution if , since in that case .
But if and , there will be at least one solution. In fact, provided and , there will be possible solutions from to .
Moreover, for each integer by which either or is less than , the number of solutions is reduced from by .
Thus, the number of solutions for is:
, where is the greater of the two integers and .
Now consider .
If and for all , then the number of solutions is at a maximum, since the may then have any values in the required range, . This maximum is , which we may prove as follows:
Suppose has the value . Then , and there are then possible pairs of values of and . Now since , takes values from to . Thus the total number of solutions is:
which is an AP with sum
(Incidentally, this is clearly , the number of ways of selecting items from , but at present I can't see why this should be so! Perhaps you can?)
So, taking the same line as when , I think the solution to is:
which we may write using sigma notation as:
For the general case, , for the maximum number of solutions and ; and, again, for every integer value below that a falls, the number of solutions will be reduced by (I think!). How do you get the maximum number of solutions? Is it ? I don't know. Perhaps you can take it a bit further, but I must sign off for the present.
PS I've now worked out the maximum number of solutions for , and indeed it is , suggesting even more strongly that my guess for the general case is correct. But I still can't see why this should be so.
The other thing to note is that in working out the solution for , I used the solution for as follows:
If , then , and the number of ways in which this can occur is the same as the problem for , with being replaced by . Could a recursive method like this be applied to the general case, or is there a more direct way?