We have $2$ $K_n$ graphs sharing exactly one vertex. Prove that no matter how we label the vertices, there are $2$ edges with the same mark 
Let $A = \{a_1, a_2, · · · , a_n\}$ and $B = \{b_1, b_2 · · · , b_n\}$ be two positive integer sets
and $|A∩B| = 1$. $C =\{$all the $2$-element subsets of $A\}∪\{$all the $2$-element subsets of $
B\}$.
Function $f : A∪B →\{0, 1, 2, · · · 2 {n \choose 2} \}$ is injective. For any ${x, y} ∈ C$, denote
$|f(x) − f(y)|$ as the mark of ${x, y}$. If $n ≥ 6$, prove that at least two elements
in $C$ have the same mark.

My idea here was to assume that all pairs from $C$ have different marks and then we easily see that every mark must appear exactly once for all pairs from $C$. From here I think that case bashing would work, but however, I am unable to do it because it seems to be a little bit too much work. Maybe there is a much more elegant solution that I am not seeing, or at least a nice way to treat all of the cases.
 A: Let $g(x) = x^{f(a_1)} + \cdots + x^{f(a_n)}, h(x) = x^{f(b_1)} + \cdots + x^{f(b_n)}$. Then we must have
$$g(x)g(x^{-1}) + h(x)h(x^{-1}) = \sum_{m = -2\binom{n}{2}}^{2\binom{n}{2}} x^m + 2n - 1.$$
In particular, if $x$ is a complex number on the unit circle, then LHS must be non-negative. If we let $N = 2\binom{n}{2}$, and $x = e^{i\theta}$, then the sum
$$\sum_{m = -N}^N x^m = \frac{x^{2N + 1} - 1}{x - 1}x^{-N} = \frac{\sin (2N + 1)\theta / 2}{\sin \theta /2}.$$
So if we let $\theta = \frac{3\pi}{(2N + 1)}$, then
$$\sum_{m = -N}^N x^m \leq -\frac{2(2N + 1)}{3\pi}.$$
So we must have
$$-\frac{2(2N + 1)}{3\pi} + (2n - 1) \geq 0.$$
or
$$\frac{(2n - 1)^2 + 1}{3\pi} \leq 2n - 1.$$
So we get $n \leq 5$ as expected.
A: A bit more context: this question asks about whether or not a certain class of windmill $K_n^{(2)}$ are graceful (i.e., admit a graceful labeling).  See Gallian's gigantic dynamic survey (pdf; p.21).  The given result is attributed to J.-C. Bermond, A. Kotzig AND J. Turgeon, On a combinatorial problem of antennas in radioastronomy, in Combinatorics, 1978, pp. 135-149 (pdf).
However, they also prove the generalized result: $K_n^{(m)}$ is ungraceful for $m \geq 1$ and $n \geq 6$.  So this includes the complete graph $K_n$ when $n \geq 6$, and any number $m \geq 1$ of complete graphs $K_n$ with $n \geq 6$ glued at a point.
It's also worth noting that $K_3^{(2)}$, $K_4^{(2)}$, and $K_5^{(2)}$ are also ungraceful (mentioned by Graham and Sloane, On additive bases and harmonious graphs SIAM J. Disc. Meth. 1980 (pdf)).  This means the claim in the question holds for $n \geq 3$ (not just for $n \geq 6$).  Of course, $K_2^{(2)}$ is a 2-edge path, and is graceful (as are all stars $K_2^{(m)}$ for $m \geq 1$).
I asked about graceful labellings of $K_n^{(k)}$ for $k \geq 3$ at MathOverflow.
