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For $K=M\cup N$, the simplicial Mayer--Vietoris sequence is

Construct a simplicial mapping cone of a degree-$k$ map of a circle. Explicitly, let the target circle have vertices $v_0,v_1,v_2$, and let the source circle have vertices $w_0,\dots ,w_{3k-1}$. Map $w_j$ to $v_{j\mathrm{mod}3}$, triangulate each quadrilateral of its mapping cylinder, and cone the source circle to one new vertex. The resulting finite simplicial complex is the mapping cone $K_k$.

Its reduced cellular, or equivalently simplicial, chain complex collapses toin degrees two and one. ConsequentlyThe same computation follows from Mayer--Vietoris applied to the cone and the mapping-cylinder neighborhood.

For a nonzero ideal $I\subset O_K$, its norm isFor $d=[K:\mathbb{Q}]$ and embeddings ${\sigma }_1,\dots ,{\sigma }_d$, defineThe field discriminant is the discriminant of any integral basis of $O_K$. If ${\alpha }_1,\dots ,{\alpha }_d$ is a $\mathbb{Z}$-basis of $I$, then

Multiplication by $\alpha$ on an integral basis has integer matrix $M_{\alpha }$. Its image lattice is $\alpha O_K$, soThe determinant of this $\mathbb{Q}$-linear map is the field norm $N_{K/\mathbb{Q}}(\alpha )$, proving

The Möbius function isMöbius inversion says thatNowis the sum of $(-1{)}^{|S|}$ over subsets of the primes whose $k$th powers divide $n$. It is one if there are no such primes and zero otherwise, proving the power-free criterion.

LetGrouping all $n$th roots of unity by their exact order givesMöbius inversion therefore yields $c_n=\mu (n)$.

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.