+1694 Find the number of ordered pairs $(a,b)$ of integers such that
\frac{a + 2}{a + 1} = \frac{b}{3}.
+16 We can do some rearrangements to do the following:
\(\frac{a + 2}{a + 1} = \frac{b}{3}\),
\(\frac{3a + 6}{a + 1} = b\),
\(\frac{3a + 3}{a + 1} + \frac{3}{a + 1} = b\)
\(3 + \frac{3}{a + 1} = b\).
From this, we have to find values of a for which \(\frac{3}{a + 1}\) is an integer. Since there's no limitation on positive or negative integers, we find that if
\(a = 2 \rightarrow b = 4\)
\(a = -4 \rightarrow b = 2\)
