jmc

Let the smaller of the two numbers be $x$. Then, the problem states that $(x+1)+x<100$. $(x+1{)}^2-x^2=x^2+2x+1-x^2=2x+1$. $2x+1$ is obviously odd, so only answer choices C and E need to be considered. $2x+1=131$ contradicts the fact that $2x+1<100$, so the answer is $\boxed{79}$