The product of two consecutive odd integers is 99. Find the integers.

We can first set the smaller one to be $a$, and the larger one to be $a+2$. Now, the problem tells us that $a(a+2)=99$. Multiplying out and rearranging, we obtain the equation $a^2+2a-99=0$. We can factor this as $(a-9)(a+11)=0$, so the solutions are $a=-11$ or $a=9$. This means that the odd integers could either be $\boxed{-11,-9}$, or they could be $\boxed{9,11}$.

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