Question

Show that $\mathrm{p}-1$is a factor of${\mathrm{p}}^{10}-1$and also of${\mathrm{p}}^{11}-1.$

Answer 1

To prove : $\mathrm{p}-1$is a factor of${\mathrm{p}}^{10}-1$and also of ${\mathrm{p}}^{11}-1$
For $\mathrm{p}-1$ to be factor of ${\mathrm{p}}^{10}-1$ and ${\mathrm{p}}^{11}-1$ the equation on substituting $\mathrm{p}=1$ should equate to$0$.

Let $\mathrm{f}\left(\mathrm{p}\right)={\mathrm{p}}^{10}-1$

⇒ $\mathrm{f}\left(1\right)={\left(1\right)}^{10}-1$

or $\mathrm{f}\left(1\right)=1-1$

$\therefore \mathrm{f}\left(1\right)=0$

Let $\mathrm{f}\left(\mathrm{p}\right)={\mathrm{p}}^{11}-1$
⇒ $\mathrm{f}\left(1\right)={\left(1\right)}^{11}-1$

or $\mathrm{f}\left(1\right)=1-1$

$\therefore \mathrm{f}\left(1\right)=0$
Based on the above two results it is clear that ${\mathrm{p}}^{10}-1$ and ${\mathrm{p}}^{11}-1$ both are completely divisible by $\mathrm{p}-1$.

Thus, $\mathrm{p}-1$is a factor of${\mathrm{p}}^{10}-1$and also of ${\mathrm{p}}^{11}-1$.

Hence Proved