The difference between the squares of two consecutive integers is 199. The sum of the squares of these two consecutive integers is
(A) 19801
(B) 39601
(C) 19602
(D) 20201
(E) 19405
Suppose that these two integers are $x$ and $x+1$, since they are consecutive.

Then $(x+1)^{2}-x^{2}=199$ or $\left(x^{2}+2 x+1\right)-x^{2}=199$ or $2 x+1=199$ or $x=99$.

Therefore, the two integers are 99 and 100, and the sum of their squares is $99^{2}+100^{2}$ or $9801+10000=19801$.

ANSWER: (A)