 Solution (source code)

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= Solution

If <Sylow subgroups> $S,T$ share $t\ne1$, they are abelian by the preceding deduction, so $S,T\leq C_G(t)$. But $|C_G(t)|=8$ and both <subgroups> have order eight. Thus $S=C_G(t)=T$. Consequently \b[distinct Sylow $2$-<subgroups> intersect only in the identity].

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