Let $Γ$ be a graph cospectral with the icosahedral graph having spectrum $\{[5]^1,[\sqrt{5}]^3, [-1]^5,[-\sqrt{5}]^3\}$. I have shown that Γ has 12 vertices, 30 edges, regular with each vertex having degree 5 and the number of common vertices between two non-adjacent vertices is less than or equal to two.
Now consider any arbitrary vertex $γ$ of $Γ$ and let $Δ$ be the subgraph of Γ induced by vertices at distance at least 2 from $γ$. It is easy to check that $\Delta$ has 6 vertices. However I need to show that $\Delta$ has 10 edges. How can this be done?
Let $A$ be the adjacency matrix of $\Gamma$.
The Hoffman polynomial of your graph is \begin{align} p(x)&= \frac{12}{(5-\sqrt{5})(5-(-1))(5-(-\sqrt{5})}(x-\sqrt{5})(x-(-1))(x-(-\sqrt{5}) \\&= \frac{1}{10}(x^2-5)(x+1).\end{align}
The main point here is that $p(x)$ is such that $P(A)=J$, where $J$ is the $12\times 12$ matrix with every entry equal to $1$. Hence $A^3=10J-A^2+5A+5I.$ Since $\Gamma$ is $5$-regular, we find that the diagonal entries of $A^3$ are all equal to $10$.
On the other hand, for a vertex $\gamma$, the $(\gamma,\gamma)$ entry of $A^3$ is exactly twice the number of triangles containing $\gamma$. We conclude that every vertex $\gamma$ is in exactly $5$ triangles.
A simple edge counting argument then reveals that $\Delta$ has exactly $10$ edges.