Split $\{1,...,3n\}$ into triples with $x+y=5z$ - no solutions? Following on from Split $\{1,2,...,3n\}$ into triples with $x+y=4z$  and For which $n\in\Bbb N$ can we divide $\{1,2,3,...,3n\}$ into $n$ subsets each with $3$ elements such that in each subset $\{x,y,z\}$ we have $x+y=3z$?
I have a rickety proof there is no way to split the numbers $\{1,...,3n\}$ into triples $(x,y,z)$ with $x+y=5z$. Perhaps someone could fix it, find their own proof or disprove it.
Thanks to Rob Pratt, $(2,3,1)$ is a solution.  Are there any others?
For the moment, ignore the need to keep all numbers between $1$ and $3N$.  Suppose we have a set of $N$ disjoint triples. Let $M$ be the highest $z$.
I thought that the smallest possible sum of $z$ would come when none of the $z$ was $M/5$ or less.  Suppose a triple is $(x_0,y_0,z_0)$ with $z_0\le M/5$.  Then both $x_0$ and $y_0$ are less than $M$.  Replace the two triples whose $z$ values are $z_0$ and $M$, with others whose $z$ values are $x_0$ and $y_0$. (This is the bit that might not be possible.)
A lower sum of $z$ would result because $x_0+y_0=5z_0\le M\lt M+z_0$.
There must be at least $N$ values for the $z$ between $M/5+1$ and $M$ so $M\ge5N/4$.  The sum of the $z$ is at least the sum of the numbers from $(N/4)+1$ to $5N/4$, which is $(3N^2+2N)/4$.
But the sum of all $3N$ numbers is $3N(3N+1)/2$, which must be exactly $6$ times the sum of the $z$.  The lowest possible sum of $z$ is too high, so there is no solution.
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 A: Disproof: $n=1$, $(x,y,z)=(2,3,1)$.
A: Modulo $6$ you need $\frac{3n(3n+1)}2$ divisible by $6.$ It is obviously divisible by $3,$ but it is divisible by $2$ only when $n\equiv 0,1\pmod 4.$
(This is because if $x+y=5z,$ then $x+y+z\equiv 0\pmod 6.$)
Any $m\geq \frac{6n}5$ must be used as an $x$ or $y,$ because no $x+y=5m\geq 6n.$

Assume we have a solution.
Let $S_{xy}$ be all the values used as $x$ or $y$ values, and $S_z$ are all the values used as $z,$ then $$\sum_{m\in S_{xy}} m=5\sum_{m\in S_z} m$$
This means if $T\subseteq S_{xy}$ you must have $$\sum_{m\in T} m\leq 5\sum_{m\notin T} m$$
Or, adding $\sum_{m\notin T}m$ to both sides:
$$\sum_{m=1}^{3n} m\leq 6 \sum_{m\notin T} m$$
But by the last note above, $T=\left\{m\mid m\geq \frac{6n}{5}\right\}$ is such a set.
So you need:
$$\sum_{k=1}^{3n} m\leq 6\sum_{k=1}^{\lceil 6n/5\rceil -1} m$$
Or:
$$\begin{align}9n^2+3n&=3n(3n+1)\\&\leq 6\lceil 6n/5\rceil (\lceil 6n/5\rceil -1)\\&\leq 6\cdot (6n/5)(6n/5+1)\quad(*)\\&=\frac{216}{25}n^2+\frac{36}{5}n \end{align}$$
(When we know $n$ specifically, we can do better than the (*) step, just by calculating $\lceil 6n/5\rceil.$ We’ll need to do to eliminate a few cases.)
Subtracting, you want $$\frac9{25}n^2\leq \frac{21}{5}n$$ or $$n\leq \frac{35}{3}.$$
So you’ve cut down to the cases $n\leq11.$
We’ve already eliminated $n=2,3,6,7,10,11$ by the modulo $4$ criterio.
We know $n=1$ is a counterexample.
We are left with $n=4,5,8,9.$ We can do better by skipping the (*) step above, and using exact value of $\lceil 6n/5\rceil$ for these $n.$ When $n=4,5,8,9$ we get $5,6,10,11,$ respectively.
Then $$\begin{align}9\cdot 4^2+3\cdot 4=156&>120=6 \cdot 5\cdot 4\\
9\cdot 5^2+3\cdot5=240&>180=6\cdot6\cdot 5\\
9\cdot 8^2+3\cdot 8=600&>540=6\cdot 10\cdot 9\\
9\cdot 9^2+3\cdot 9=756&>660=6\cdot 11\cdot 10\end{align}$$
So there is no solution other than $n=1.$
