smc

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 $(A{x}^2+Bx+C{)}^2$ for some $(A,B,C)$. Setting $(A{x}^2+Bx+C{)}^2=x^4+2x^3+3x^2+2x+1$, we can compare coefficients. From the $x^4$ coefficient, we get $A=\pm 1$. Note that if $(A{x}^2+B{x}^2+C{)}^2$ works, so does $(-A{x}^2-Bx-C{)}^2$, 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 $x\equiv 0(\mathrm{mod}2)$, then $x^2\equiv 0^2(\mathrm{mod}2)$. This means $x^2+x+1\equiv 0+0+1\equiv 1(\mathrm{mod}2)$. If $x\equiv 1(\mathrm{mod}2)$, then $x^2\equiv 1^2(\mathrm{mod}2)$, and $x^2+x+1\equiv 1+1+1\equiv 1(\mathrm{mod}2)$. Either way, $x^2+x+1$ is an odd integer, and the answer is $\boxed{\text{always an odd integer}}$