14. The sides, in $\mathrm{cm}$, of two squares are integers. The difference between the areas of the two squares is $19 \mathrm{c} \mathrm{m} \mathrm{1}$. What is the sum of the perimeters of the two squares?
A $38 \mathrm{~cm}$
B $60 \mathrm{~cm}$
C $64 \mathrm{~cm}$
D $72 \mathrm{~cm}$
E $76 \mathrm{~cm}$
SolUtion
E

Let the side-length of the two squares, in $\mathrm{cm}$, be $x$ and $y$. Because the difference between their areas is $19 \mathrm{~cm}^{2}, x^{2}-y^{2}=19$. That is, $(x+y)(x-y)=19$. Since $x+y$ and $x-y$ are positive integers, it follows that $x+y=19$.

It follows that the sum of the perimeters is $4 x+4 y \mathrm{~cm}$, which is $4(x+y) \mathrm{cm}=(4 \times 19) \mathrm{cm}$, which is $76 \mathrm{~cm}$.