Mixer_V wrote:

I did some simulation. For number of cards as odd integers, the solution is simple. either all 0 or $n-1$. For even numbers the solutions are more complicated

After play around the code, I notice (not yet to prove) that besides all 0 and all 16, the other solutions (if exist) are of the form $(a,a,a,..b,b,b,b)$, where $a + b +1 =0$. So for 17 cards, we have two extra solutions: 5 pieces of $-6$ + 12 pieces $5$ 5 pieces of $-2$ + 12 pieces $1$

Assume $1 \leq i \leq 17$. Let $x_i$ be the integer written on the card $i$. Futhermore let $(1) \;\; S_k = x_1^k + x_2^k + ... + x_{17}^k, k \in \{1,2\}$. The given condition yields $x_i^2 = S_1 - x_i$, i.e. $(2) \;\; x_i^2 + x_i = S_1$. The fact that $x_i=a$ is a solution of equation (2) iff $x_i = -1 - a$ is a solution of equation (2) implies $x_i=x_1$ or $x_i=-1-x_1$. Let $n$ be the number of $i$´s for which $x_i = -1 - x_1$. According to equation (2) $\sum_{i=1}^{17} (x_i^2 + x_i) = \sum_{i=1}^{17} S_1$ $S_2 + S_1 = 17S_1$ $S_2 = 16S_1$ $(17 - n)x_1^2 + n(-1 - x_1)^2= 16[(17 - n)x_1 + n(-1 - x_1)]$ $n(2x_1 + 1) = x_1(16 - x_1)$, yielding $(3) \;\; n = \frac{1}{4} \Big( -2x_1 + 33 - \frac{33}{2x_1+1} \Big)$. Hence $2x_1 + 1 \mid 3 \cdot 11$, which give us $(x_1,n) = (-17,17), (-6,12), (-2,12), (-1,17), (0,0), (1,5), (5,5), (16,0)$. Now $n \neq 17$ since $n < 17$, which give us the following four solutions: $\bullet$ All 17 cards have the same number $m$, where $m \in \{0,16\}$. $\bullet$ 5 cards have the number -2 and 12 cards have the number 1. $\bullet$ 5 cards have the number -6 and 12 cards have the number 5.

Solar Plexsus wrote: Assume $1 \leq i \leq 17$. Let $x_i$ be the integer written on the card $i$. Futhermore let $(1) \;\; S_k = x_1^k + x_2^k + ... + x_{17}^k, k \in \{1,2\}$. The given condition yields $x_i^2 = S_1 - x_i$, i.e. $(2) \;\; x_i^2 + x_i = S_1$. The fact that $x_i=a$ is a solution of equation (2) iff $x_i = -1 - a$ is a solution of equation (2) implies $x_i=x_1$ or $x_i=1-x_1$. Let $n$ be the number of $i$´s for which $x_i = 1 - x_1$. According to equation (2) $\sum_{i=1}^{17} (x_i^2 + x_i) = \sum_{i=1}^{17} S_1$ $S_2 + S_1 = 17S_1$ $S_2 = 16S_1$ $(17 - n)x_1^2 + n(1 - x_1)^2= 16[(17 - n)x_1 + n(1 - x_1)]$ which is equivalent to $(3) \;\; 15n(2x_1 - 1) = -17x_1(x_1 - 16)$. Assume $17 \nmid 2x_1 - 1$. Then $17|n$ by equation (3), which give us $n=0$ (since $n<17$). Hence $x_i=x_1$ for all $1 \leq i \leq 17$, where $x_1=0$ or $x_1=16$ by equation (3). Next assume $17 \mid 2x_1 - 1$. The fact that $GCD(x_1,2x_1-1)=1$ combined with equation (3) implies $\frac{34(x_1 - 16)}{2x_1 - 1} = 17 - \frac{17 \cdot 31}{2x_1 - 1} \in \mathbb{Z}$, which (since $17 \mid 2x_1-1$) means $(4) \;\; 2x_1 - 1 = -17t, t \in \{-31,-1,1,31\}$. Then by equation (4) $2x_1 - 1 \equiv -17 \cdot (\pm 1) = \mp 17 \equiv \mp 2 \pmod{5}$, yielding $(5) \;\; x_1 \equiv 2,4 \pmod{5}$. But $x_1 \equiv 0,1 \pmod{5}$ by equation (3), contradicting congruence (5). Conclusion: The integers written on all 17 cards are the same. The number is either 0 or 16. but 5 pieces of $-6$ + 12 pieces of $5$ 5 pieces of $-2$ + 12 pieces of $1$ are indeed two valid solutions. You can verify them by hand.

I have edited my solution. My first solution was wrong because I obtain $x_i = 1 - x_1$ instead of $x_i = -1 - x_1$ (which is correct)