+1839 Find the number of ordered pairs $(a,b)$ of integers such that
\frac{a + 2}{a + 1} = \frac{b}{8}.
+1768 Multiplying both sides by 8(a+1) (which is nonzero since a is an integer), we get: $8(a+2)=b(a+1).$This forces b to be a multiple of 8. We can rewrite the equation as: $b=8k⋅a+1a+2=8k(1+a+11)$Since b is an integer, k must also be an integer.
If a=−1, then the fraction becomes undefined, so we must reject this possibility. Otherwise, a+11 is an integer, so a+1 must divide 1. The only two possibilities are a=0 and a=−2.
If a=0, then b=16k, which means that k can be any integer. So, in this case, there are infinitely many solutions (a,b).
If a=−2, then b=−8k, which means that k must be nonpositive. The only nonpositive integer that divides 8 is −8, so k=−8 and b=64. Thus, the only solution in this case is (−2,64).
In conclusion, there are ∞ solutions if a=0 and exactly 1 solution if a=−2.
