Let $U_k=(1+kd)^2$. Then $U_{k+3}-U_{k+2} -U_{k+1}+U_k=4d^2$, 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 $(1+d)^2+(1+2d)^2 +\cdots +(1+Nd)^2$, 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 $(1+kd)^2$ to a minus sign, we decrease the sum by $2(1+kd)^2$. In particular, if $k \equiv 0 \pmod {2d}$, we decrease the sum by 2 (modulo $4d^2$). 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.
There is not much of a complication if instead of $1+kd$ we use $a+kd$, where $a$ and $d$ are relatively prime.