, 2011

Let $U_k = (1 + k d)^2$. We calculate $U_{k+3} - U_{k+2} - U_{k+1} + U_k$. This turns out to be $4d^2$, a constant. Changing signs, we obtain the sum $-4d^2$.
Thus if we have found an expression for a certain number $S_0$ as a sum of the desired type, we can obtain an expression of the desired type for $S_0 + (4d^2)q$, for any integer $q$.

It remains to show that for any $S$, there exists an integer $S'$ such that $S' \equiv S \pmod{4d^2}$ and $S'$ can be expressed in the desired form. Look at the sum
$$(1+d)^2 + (1+2d)^2 + \dots + (1+Nd)^2,$$
where $N$ is "large." We can at will choose $N$ so that the sum is odd, or so that the sum is even.
By changing the sign in front of $(1 + k d)^2$ to a minus sign, we decrease the sum by $2(1 + k d)^2$. In particular, if $k \equiv 0 \pmod{2d}$, we decrease the sum by $2$ (modulo $4d^2$). So
If $N$ is large enough, there are many $k < N$ such that $k$ is a multiple of $2d$. By switching the sign in front of $r$ of these, we change ("downward") the congruence class modulo $4d^2$ by $2r$. By choosing $N$ so that the original sum is odd, and choosing suitable $r < 2d^2$, we can obtain numbers congruent to all odd numbers modulo $4d^2$. By choosing $N$ so that the original sum is even, we can obtain numbers congruent to all even numbers modulo $4d^2$. This completes the proof. $\square$