Find all pairs (m,n) of positive integers such that m(n+1) + n(m-1) = 2000

To find all pairs (m, n) of positive integers that satisfy the equation m(n+1) + n(m-1) = 2000, we can use a systematic method of exploration.

Let's rewrite the equation as follows:

mn + m + mn - n = 2000
2mn + m - n = 2000

To simplify the equation further, we can divide both sides by 2:

mn + (m - n)/2 = 1000

Now, let's consider the possibilities for the value of m - n:

- If m > n, then (m - n)/2 is a positive integer.
- If n > m, then (m - n)/2 is a negative integer.

We can start by assuming that m > n and proceed with this assumption.

Let's list possible values for (m - n)/2 such that (m - n) is positive:

m - n = 2, 4, 6, 8, ..., 1998

Now, re-write the equation with the assumption that m - n can take the above values:

mn + (m - n)/2 = 1000

Substitute (m - n) with the above values and solve for m and n:

When m - n = 2:
2n + 1 = 1000
2n = 999
n = 499.5

Since n should be a positive integer, we discard this solution.

Continue this process for each possible value of (m - n) until we reach a suitable solution.

The process described above will allow you to determine all pairs of positive integers (m, n) that satisfy the given equation.