Well, to find a satisfactory $b$, we can make use of our knowledge of the decomposition $45=3^2\times 5$.

This means if we pick $b=3\times5=15$, then we would have $b^n\equiv0\pmod{45}$ for any $n\ge 2$ (in particular for $n=45$).

Well, to find a satisfactory $b$, we can make use of our knowledge of the decomposition $45=3^2\times 5$.

This means if we pick $b=3\times5=15$, then we would have $b^n\equiv0\pmod{45}$ for any $n\ge 2$ (in particular for $n=45$).

[Added for clarity]

More explicitly, note that $15^2=3^2\times5^2=45\times 5$. Hence if $n\ge2$, then $$15^n\equiv 15^2\times15^{n-2}\equiv 45\times 5\times 15^{n-2}\equiv 0\pmod {45}$$

Perhaps making use of the known factorization is in some sense cheating. In that case, you can just say $15^2=225=5\times 45$, and so $15^2\equiv 0\pmod{45}$, and then use $$15^n\equiv 15^2\times15^{n-2}\equiv 0\times15^{n-2}\equiv0\pmod{45}$$

However, our motivation for looking at the number $15$ does come from us already knowing the factorization of $45$.

Well, to find a satisfactory $b$, we can use the fact that we already know a decomposition of $45=3^2\times 5$. This means if we pick $b=3\times5=15$, then we would have $b^n\equiv0\pmod{45}$ for any $n\ge 2$ (in particular for $n=45$).

Well, to find a satisfactory $b$, we can make use of our knowledge of the decomposition $45=3^2\times 5$. 

This means if we pick $b=3\times5=15$, then we would have $b^n\equiv0\pmod{45}$ for any $n\ge 2$ (in particular for $n=45$).