Let's ask ourselves the question: How many pieces of 56 grams can we possibly have if we want a 17-piece solution? This will help us limit the groupings in the 9-way split.
Let's assume we have three pieces of 56 grams. Then in the 8-way split scenario, one of the 56-gram pieces must be in a 2-piece group to get 63 grams, and thus there must exist (at least) two pieces of 7 grams. We also know that there is either a third piece $K$ of 7 grams, or exactly two pieces $K_1, K_2$ which add up to 7 grams.
In the 7-way split scenario, each 63-gram group needs to have at least two pieces of gold (otherwise, there will be a single piece of 72 grams > 56 grams). This means that we have 3 extra pieces of gold to assign to groups in the 7-way split. The known two pieces that weigh 7 grams cannot be in a group with only two pieces (otherwise the other piece in the group would be heavier than 56 grams). So the following three possibilities exist:
One group only with more than 2 pieces:
 + [ 7] + [ 7] + [A?] + [B?]
We know this cannot be true, since $K$ or both $K_1, K_2$ must be in this group, making the total larger than 72 grams.
Two groups with more than 2 pieces, a 3-piece group and a 4-piece group:
 + [ 7] + [A?] + [C1?]
 + [B?] + [C2?]
 + 
Because of the pigeonhole principle, we cannot have $K = 7$ be in one of the these groups (which would make at least one group unequal to 72 grams total), nor A = $K_1$, B = $K_2$ (for the same reason). Thus we have C1 = $K_1$, C2 = $K_2$. The last 7-gram piece also needs to be in one of these groups, so either:
B = 7 $\Rightarrow$ $K_2$ = 9 (contradiction)
A = 7 $\Rightarrow$ $K_1$ = 2 $\Rightarrow$ $K_2$ = 5 $\Rightarrow$ B = 11
We will now do some Sudoku-like deductions and reach a contradiction. The 11-gram piece will occupy a 2-piece group in the 8-way split, and thus generate a 52-gram piece, which will generate a 4-gram piece in the 9-way split, and this 4-gram piece must be part of a 2-piece group in the 8-way or 7-way split, and generate a piece that is larger than 56 grams.
Three groups with more than 2 pieces, 3 pieces each:
 + [ 7] + [B1?]
 + [ 7] + [B2?]
 + [A?] + [B3?]
We also know this cannot be true, because either $K_1, K_2$ must be in this group (which by the pigeonhole principle would make at least one group unequal to 72 grams total), or A = $K$, and $B_i = 9, \forall i$.
In this latter case, the pigeonhole principle again requires at least one $B_i$ to be in a two-piece group in the 8-way split and generate a 54-gram piece, and at least one 7-gram and one 9-gram piece to be in a two-piece group in the 9-way split and generate a 49-gram and a 47-gram piece, respectively. The 54-gram, 49-gram and 47-gram pieces must then generate 18-gram, 23-gram and 25-gram pieces in the 7-way split, leaving two free pieces that must add up to 72. But since the 54-piece must either create two pieces of 1 grams or one piece of 2 grams in the 9-way split, we reach a contradiction.
From the above, it is clear that there can be at most two 56-gram pieces in a 17-piece solution. This implies that the 9-way split has at least six 2-piece groups, plus either a 56-gram single-piece group + one 3-piece group (two 56-gram pieces case), or two 2-piece groups (single 56-gram piece case).
We will now show that the single 56-gram piece case admits no solutions with 17 pieces. To do this, we will consider the even/odd parity required for each of the groups in the 8-way split. There are 7 2-piece groups and one 3-piece groups in the 8-way split. Since each group must sum to 63 grams (odd), each 2-piece group must have one odd-weighted piece and one even-weighted piece. (The 3-piece group is either one odd or all odd pieces.)
Correspondingly, the 9-way split has 8 groups of 2 pieces, and one group of a single 56-gram piece. These 2-piece groups must be either all odd-weighted or all even weighted. Therefore, any pairing in the 8-way split cannot appear in the 9-way split. Moreover, this means that if we fix the location and the weight of any two pieces in the 8-way split not in the same group, the weights of all the other pieces can be determined, relative to the fixed weights.
For example, consider the scenario in which the 56-gram piece is in the 2-piece group, which generates a piece of 7 grams in the same group, which, when put in a (2-piece) group in the 9-way split, generates a piece of 49 grams, which can then be put back in the 8-way split again, so on and so forth.
I did this for all the possible scenarios with a single 56-gram piece in the 8-way split:
- 56-gram piece in 2-piece group $\Rightarrow$ 7-gram weight $\Rightarrow$ 49-gram in 9-way split $\Rightarrow$ 49-gram put into 2-piece group in 8-way split ...
- 56-gram piece in 2-piece group $\Rightarrow$ 7-gram weight $\Rightarrow$ 49-gram in 9-way split $\Rightarrow$ 49-gram put into 3-piece group (all odd / one odd only) in 8-way split ...
- 56-gram piece in 3-piece group $\Rightarrow$ $K_1 = 7 - K_2, K_2$ $\Rightarrow$ $7-K_2$ put into 9-way split ...
In all these scenarios, a full, self-consistent set of weights will be generated, all dependent on a single variable, in the form of $n_i-X$ or $n_i+X$, where $X$ is the independent variable. One example is as follows:
This is for scenario 3 above, with independent variable $X = K_2$. It is easy to see that no two pieces with the same weight parity can sum to 72 grams if they have opposing signs (such that the independent variable $X$ is cancelled out). This is because for all scenarios, $n_i$ must be a multiple of 7 between 0 and 56, and even and odd pairs can never share $n_i$ by construction. (I know "by construction" is not rigorous at all, but I don't know of a clever way to prove this more elegantly.)
The only possibility is if the signs are the same, in which case the value of X will be forced to one value. However, this means that only a maximum of 1 same-parity pair can sum to 72 grams. By the pigeonhole principle, there are at least 4 2-piece groups in the 7-way split scenario, one of which is guaranteed to be both odd in weight parity, and the other guaranteed to be both even in weight parity. Therefore, the 7-way split has at least one 2-piece group that can never sum to 72 grams, and our proof for the single 56-gram case is done.
We are now left with the case where exactly two 56-gram pieces (and one 7-gram piece) exist in the 17-piece solution. If we can prove this case hsa no valid solutions, we are done. However, I was lazy, so I just brute-forced all possible combinations. I found no valid solutions in this case, and so the proof is done.