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 \(\boxed{9,11}\).
Hope this helped 
