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.