Sweet sixteen

Case 1: The same number

Let $x$ be the number repeated in the collection. Since the square of one number is equal to the sum of the rest of the collection, $x^2 = 16x$ , which solves to $x = 0$ or $x = 16$ . Ignoring $x = 0$ as the trivial case of seventeen $0$ 's, this leaves one collection of seventeen $16$ 's.

Case 2: Two different numbers

Let the two different numbers in the collection be $x$ and $y$ such that $x \neq y$ , and let there be $m$ $x$ 's and $n$ $y$ 's. Since the square of one number is equal to the sum of the rest of the collection, $x^2 = mx + ny - x$ and $y^2 = mx + ny - y$ , so $mx + ny = x^2 + x = y^2 + y$ , which means either $y = x$ or $y = -x - 1$ . However, since the numbers in this case must be different, $x \neq y$ , so $y = -x - 1$ . Since there are $17$ total numbers, $m + n = 17$ . These equations combine to $m = \frac{(x + 17)(x + 1)}{(2x + 1)}$ , which have integer solutions $(x, m)$ at $(1, 12)$ and $(5, 12)$ . $(1, 12)$ corresponds with the collection of twelve $1$ 's and five $-2$ 's, and $(5, 12)$ corresponds with the collection of twelve $5$ 's and five $-6$ 's.

Case 3: Three (or more) different numbers

Let the three different numbers in the collection be $x$ , $y$ , and $z$ such that $x \neq y \neq z$ , and let there be $m$ $x$ 's, $n$ $y$ 's, and $p$ $z$ 's. Since the square of one number is equal to the sum of the rest of the collection, $x^2 = mx + ny + pz - x$ , $y^2 = mx + ny + pz - y$ , and $z^2 = mx + ny + pz - z$ , so $mx + ny + pz = x^2 + x = y^2 + y = z^2 + z$ , which means either $y = x$ or $y = -x - 1$ , $z = y$ or $z = -y - 1$ , and $z = x$ or $z = -x - 1$ . However, since the numbers in this case must be different, $x \neq y \neq z$ , so $y = -x - 1$ , $z = -y - 1$ , $z = -x - 1$ , but then these equations lead to $x = y$ which is also not true. Therefore, there are no collections with three (or more different numbers).

Summary

The different collections include seventeen $16$ 's, twelve $1$ 's and five $-2$ 's, and twelve $5$ 's and five $-6$ 's, for a total of $\boxed{3}$ different collections.

For any number $x$ in the collection, it must be the case that $x^2 = S-x,$ where $S$ is the sum of the collection. So $x^2+x = S$ for all $x$ in the collection. This equation has at most two roots, so the collection consists of at most two distinct integers, and if there are two distinct integers, they sum to $-1.$

That is, there are $k$ copies of $a-1,$ and $17-k$ copies of $-a,$ for some integer $a$ and some integer $k$ with $0 \le k \le 17.$

Here $S = 2ka - 17a - k,$ and so $a^2-a = 2ka-17a-k,$ or $a^2 + (16-2k)a + k = 0.$ Quickly checking all the integer values of $k$ between $0$ and $17$ for which this equation has integer roots, we get $k = 0, 5, 12, 17.$

If $k = 0,$ we get $a = 0, 16,$ which leads to either 17 copies of 0 or 17 copies of 16. If $k=17,$ we get $a = 1, 17,$ which leads to 17 copies of 0 or 17 copies of 16.

If $k=5,$ we get $a = -1, -5,$ which leads to 5 copies of -2 and 12 copies of 1, or 5 copies of -6 and 12 copies of 5. If $k=12,$ we get $a=2,6,$ which leads to 12 copies of 1 and 5 copies of -2, or 12 copies of 5 and 5 copies of -6.

Each collection has been listed twice in this enumeration. Since 17 copies of 0 are not in our list, this leaves $\fbox{3}$ distinct collections.

Let any two of the 17 numbers be a and b.

Let the sum of the other 15 be k.

a^2=b+k b^2=a+k.

take difference a^2-b^2=b-a

Thus a=b or a+b=-1 (dividing through by a-b)

Therefore any number equals the first number or is -1 minus it. Therefore there are at most 2 distinct numbers.

Let the collection consist of n copies of the number m and 17-n copies of the number -1-m.

m^2=(n-1)m+(17-n)(-1-m)

m^2+(18-2n)m+(17-n) =0

Quadratic formula

m=n-9 +- sqrt (n^2-17n+64)

Try n=1 to 17 to find integer solutions for m

n = 17 or 5 (there is a n=12 solution but of course it is the n=5 solution, the other way around).

m = 16 or (-2 or -6) respectively

So 17 copies of 16 works

5 copies of -2 and 12 copies of 1 works

5 copies of -6 and 12 copies of 5 works

These can be verified

3 solutions