Question

Find the number of ordered pairs (m,n) of integers that satisfy
mn = 3m + 6n + 2m - 5n + 1.

Answer 1

We start with the equation given in the problem:

\[
mn = 3m + 6n + 2m - 5n + 1.
\]

First, we combine like terms on the right side:

\[
mn = (3m + 2m) + (6n - 5n) + 1,
\]

which simplifies to:

\[
mn = 5m + n + 1.
\]

Next, we rearrange the equation into a standard quadratic form:

\[
mn - 5m - n - 1 = 0.
\]

Let’s introduce \(m\) and factor it out. We can rewrite the equation as:

\[
mn - 5m - n - 1 = 0 \implies m(n - 5) - n - 1 = 0.
\]

This can be expressed as:

\[
m(n - 5) = n + 1.
\]

From this, we can solve for \(m\):

\[
m = \frac{n + 1}{n - 5}.
\]

For \(m\) to be an integer, the right-hand side must also be an integer, which implies that \(n - 5\) must divide \(n + 1\).

Now let's express \(n + 1\) in terms of \(n - 5\):

\[
n + 1 = (n - 5) + 6.
\]

Hence:

\[
m = \frac{(n - 5) + 6}{n - 5} = 1 + \frac{6}{n - 5}.
\]

For \(m\) to be an integer, \(\frac{6}{n - 5}\) must also be an integer. This means that \(n - 5\) must be a divisor of 6.

The divisors of 6 are \( \pm 1, \pm 2, \pm 3, \pm 6\). Let’s evaluate the possible values of \(n - 5\) based on these divisors:

1. \(n - 5 = 1 \implies n = 6\)
2. \(n - 5 = -1 \implies n = 4\)
3. \(n - 5 = 2 \implies n = 7\)
4. \(n - 5 = -2 \implies n = 3\)
5. \(n - 5 = 3 \implies n = 8\)
6. \(n - 5 = -3 \implies n = 2\)
7. \(n - 5 = 6 \implies n = 11\)
8. \(n - 5 = -6 \implies n = -1\)

Now we substitute each of these values back into the formula for \(m\):

- If \(n = 6\), then \(m = 1 + \frac{6}{1} = 7\).
- If \(n = 4\), then \(m = 1 + \frac{6}{-1} = -5\).
- If \(n = 7\), then \(m = 1 + \frac{6}{2} = 4\).
- If \(n = 3\), then \(m = 1 + \frac{6}{-2} = -2\).
- If \(n = 8\), then \(m = 1 + \frac{6}{3} = 3\).
- If \(n = 2\), then \(m = 1 + \frac{6}{-3} = -1\).
- If \(n = 11\), then \(m = 1 + \frac{6}{6} = 2\).
- If \(n = -1\), then \(m = 1 + \frac{6}{-6} = 0\).

So, we have the following ordered pairs \((m, n)\):

1. \((7, 6)\)
2. \((-5, 4)\)
3. \((4, 7)\)
4. \((-2, 3)\)
5. \((3, 8)\)
6. \((-1, 2)\)
7. \((2, 11)\)
8. \((0, -1)\)

Thus, we find a total of 8 ordered pairs \((m,n)\) satisfying the original equation. Therefore, the final answer is:

\[
\boxed{8}.
\]