Let $a=x, b=x+1, c = x(x+1)$.  Then $D = x^2 + (x+1)^2 + x^2(x+1)^2$, which simplifies to $x^4 + 2x^3 + 3x^2 + 2x + 1$.
From the options, we want to test if $D$ is always a perfect square.  Because the polynomial expression for $D$ is quartic, if it is a perfect square, it would be the square of a quadratic expression.  Thus, $D$ could be written in the form $(Ax^2 + Bx + C)^2$ for some $(A, B, C)$.
Setting $(Ax^2 + Bx + C)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1$, we can compare coefficients.  From the $x^4$ coefficient, we get $A = \pm 1$.  Note that if $(Ax^2 + Bx^2 + C)^2$ works, so does $(-Ax^2 - Bx - C)^2$, so we can arbitrarily pick $A=1$.
We now have $(x^2 + Bx + C)^2= x^4 + 2x^3 + 3x^2 + 2x + 1$.  Setting the cubic terms equal gives $2B = 2$, or $B = 1$.  This leaves $(x^2 + x + C)^2 =  x^4 + 2x^3 + 3x^2 + 2x + 1$.  We can quickly inspect the constant term to determine that $C = \pm 1$.  We reject $C = -1$, since the quadratic and linear terms won't match up, which leaves $(x^2 + x + 1)^2$ as the only possibility - and, in fact, it works.
Thus, $D$ is always the square of an integer - namely $x^2 + x + 1$. This in turn means that $\sqrt{D}$ is always rational, which leaves choices $A, B, C$ as the only possible correct answers.
The question now is whether $\sqrt{D}$, or $x^2 + x + 1$, is odd, even, or could be both.  We have two cases for $x$:
If $x \equiv 0 \pmod 2$, then $x^2 \equiv 0^2 \pmod 2$.  This means $x^2 + x + 1 \equiv 0 + 0 + 1 \equiv 1 \pmod 2$.  
If $x \equiv 1 \pmod 2$, then $x^2 \equiv 1^2 \pmod 2$, and $x^2 + x + 1 \equiv 1 + 1 + 1 \equiv 1 \pmod 2$.  
Either way, $x^2 + x + 1$ is an odd integer, and the answer is $\fbox{C}$
Let $D=a^2+b^2+c^2$, where $a$, $b$, are consecutive integers and $c=ab$.  Then $\sqrt{D}$ is:
$\textbf{(A)}\ \text{always an even integer}\qquad \textbf{(B)}\ \text{sometimes an odd integer, sometimes not}\\ \textbf{(C)}\ \text{always an odd integer}\qquad \textbf{(D)}\ \text{sometimes rational, sometimes not}\\ \textbf{(E)}\ \text{always irrational}$