The multinomial expansion for the power of $2$ is the key to solving this problem. The expansion of the square of the sum of elements $X_1,X_2,\dots ,X_n$ in the list $X$ looks like:

$${\displaystyle \begin{array}{rl}\text{square of sum}& =(X_1+X_2+X_3+\cdots +X_N{)}^2\\ & =X_1^2+X_2^2+X_3^2+\cdots +X_N^2+2\cdot X_1\cdot X_2+2\cdot X_2\cdot X_3+2\cdot X_1\cdot X_3+\cdots +2\cdot X_{N-1}\cdot X_N\\ & =\text{sum of squares}+2\cdot \text{sum of pairwise products}\end{array}}$$

Let $S(X)$ be the sum of elements, $SQ(X)$ be the sum of squares of elements, and $SP(X)$ be the sum of all pairwise products of elements of the list $X$. We can now rewrite the above equation as:

$${\displaystyle S(X{)}^2=SQ(X)+2\cdot SP(X)}$$

We can also observe the following about how the above values change when an additional element $n$ is added to the list $X$:

$${\displaystyle \begin{array}{rl}S(X+[n])& =S(X)+n\\ SQ(X+[n])& =SQ(X)+n^2\\ SP(X+[n])& =SP(X)+n\cdot S(X)\end{array}}$$

Our task is to achieve $S(E^{\prime }{)}^2=SQ(E^{\prime })$, where $E^{\prime }$ is the extended list that we get by adding extra elements to $E$. In other words, we want to make $SP(E^{\prime })=0$.

Test Set 1: $K=1$

If we are allowed only a single addition, we must choose an element $n$ such that $SP(E+[n])=0$.

$${\displaystyle \begin{array}{r}SP(E+[n])=0\\ ⟹SP(E)+n\cdot S(E)=0\\ ⟹n\cdot S(E)=-SP(E)\end{array}}$$

If $S(E)\ne 0$, we can get a squary list whenever $-SP(E)/S(E)$ is an integer, which happens if and only if $S(E)$ divides $SP(E)$. In that case, $-SP(E)/S(E)$ is our answer.

If $S(E)=0$, then $S(E+[n])=n$. Since we want $S(E+[n]{)}^2=SQ(E+[n])$, we need $SQ(E+[n])=n^2$. This is possible only if $SQ(E)=0$, that is, if all elements in $E$ are zeros. In this case, we can choose any value as our answer. But if any element in $E$ is not zero, it is impossible to get a squary list with only one addition.

Test Set 2: $K>1$

At first, the search space might seem hopelessly broad here. But we can observe (or surmise and then confirm) that it is always possible to get a squary list by adding only two elements:

• $n_1=1-S(E)$
• $n_2=-SP(E+[n_1])$

After adding $n_1$, we have

$${\displaystyle S(E+[n_1])=1}$$

After adding $n_2$, we have

$${\displaystyle \begin{array}{rl}SP(E+[n_1,n_2])& =SP(E+[n_1])+n_2\cdot S(E+[n_1])\\ & =SP(E+[n_1])+(-SP(E+[n_1]))\cdot 1\\ & =0\end{array}}$$

Thus, the two numbers satisfy the condition $SP(E^{\prime })=0$. We can also see that since the numbers in the original list are each of magnitude no greater than ${10}^3$, $|n_1|\le {10}^6+1$, and $|n_2|\le 2\cdot {10}^{12}$, both well within the limits.