Solution: $\mathbf{D}$
Suppose $a+12 \sqrt{2}=(x+y \sqrt{2})^{2}$, where $a, x$ and $y$ are positive integers. It follows that $a+12 \sqrt{2}=x^{2}+2 x y \sqrt{2}+2 y^{2}=\left(x^{2}+2 y^{2}\right)+2 x y \sqrt{2}$. Therefore $x y=6$ and $x^{2}+2 y^{2}=a$. Since $x$ and $y$ are positive integers, the only possibilities for $x$ and $y$ are $x=1, y=6 ; x=2, y=3 ; x=3, y=2$ and $x=6, y=1$. Therefore the only possibilities for $a$ are $1^{2}+2\left(6^{2}\right)=73 ; 2^{2}+2\left(3^{2}\right)=22$; $3^{2}+2\left(2^{2}\right)=17$ and $6^{2}+2\left(1^{2}\right)=38$. Therefore options A, B, C and E are numbers which have square roots of the required form, but not option $\mathrm{D}$.
21. Which of the following numbers does not have a square root in the form $x+y \sqrt{2}$, where $x$ and $y$ are positive integers?
A $17+12 \sqrt{2}$
B $22+12 \sqrt{2}$
C $38+12 \sqrt{2}$
D $54+12 \sqrt{2}$
E $73+12 \sqrt{2}$