= Solution
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> satisfies
$$
A^2=(k-b)I+(a-b)A+bJ.
$$
On the orthogonal complement of the all-one <vector>, the two possible <eigenvalues> are
$$
r,s=\frac{a-b\pm\sqrt{(a-b)^2+4(k-b)}}2.
$$
Using $1+m_r+m_s=n$ and $k+m_rr+m_ss=0$ 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> is
$$
3^{(1)},\qquad1^{(5)},\qquad(-2)^{(4)}.
$$
If 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 $\mathbf1^\perp$ intersect nontrivially. For a nonzero common <vector> $v$, $Av=Bv=v$ and therefore
$$
Cv=(J-I-A-B)v=-3v,
$$
contradicting the Petersen <spectrum>. No such partition exists.
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