Question

If pbe a natural number, then prove that $p^{n+1}+(p+1{)}^{2n-1}$ is divisible by $p^2$ +p + 1 for every positive integer n.

Answer 1

Let f (n) = $p^{p+1}+(p+1{)}^{2n-1}$
We have f (1) = $p^2$ + p + 1 so that f (1) is divisible by $p^2$ + p + 1.
Now assume that f (m) is divisible by $p^2$ + p + p + 1 i.e.,
we assume that
$p^{m+1}+(p+1{)}^{2m-1}=k(p^2+p+1)$
Now f (m + 1) = $p^{m+2}+p(p+1{)}^{2m+2-1}$
$=p^{m+2}++\left[k\right(p^2+p+1)-p^{m+1}](p+1{)}^2$
$=p^{m+2}-(p+1{)}^2{p}^{m+1}+k(p+1{)}^2(p^2+p+1)$
$=p^{m+1}(p-p^2-2p-1)+k(p+1{)}^2(p^2+p+,1)$
$=(p^2+p+1)\left[k\right(p+1{)}^2-p^{m+1}]$
Hence f (m + 1) is divisible by $p^2$ + p + 1.
$\therefore$ By induction, f (x) is divisible by $p^2$ + p + 1 for all n $ϵ$ N.