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The topological group $S^1=\{z\in \mathbb{C}:|z|=1\}$ has complex multiplication and the subspace topology. It is compact and abelian, so every finite-dimensional irreducible complex representation is one-dimensional. A continuous character lifts along $t\mapsto e^{it}$ to a continuous homomorphism $\mathbb{R}\to {\mathbb{C}}^{\times }$, hence has the form $t\mapsto e^{i\lambda t}$. Periodicity by $2\pi$ forces $\lambda =n\in \mathbb{Z}$, giving precisely $z\mapsto z^n$.

A strongly regular graph with parameters $(k,a,b)$ is $k$-regular, with every adjacent pair having $a$ common neighbours and every distinct nonadjacent pair having $b$ common neighbours. Its adjacency matrix satisfiesOn the orthogonal complement of the all-one vector, the two possible eigenvalues areUsing $1+m_r+m_s=n$ and $k+m_{r}r+m_{s}s=0$ gives exactly the two displayed expressions for $m_r,m_s$. They are eigenspace dimensions and hence integers, proving the rationality condition.

The Petersen graph has parameters $(3,0,1)$, so its spectrum isIf three Petersen graphs partitioned $E(K_{10})$, their adjacency matrices would satisfy $A+B+C=J-I$. The five-dimensional eigenvalue-one spaces of $A$ and $B$ inside the nine-dimensional space $1^{\perp }$ intersect nontrivially. For a nonzero common vector $v$, $Av=Bv=v$ and thereforecontradicting the Petersen spectrum. No such partition exists.