If p is a fixed positive integer, prove by induction that `p^(n +1) + (p + 1)^(2n - 1)` is divisible by `P^2+ p +1` for all `n in N`.

Explanation

`P(n) : p^(n+1) + (p+1)^(2n-1)` is divisible by `p^(2) + p + 1`
For `n = 1, P(1) : p^(2) + p + 1`
So, `P(1)` is true.
Let `P(k)` be true.
Then `p^(k+1) + (p+1)^(2k-1)` is divisible by `p^(2) + p + 1`
`rArr p^(k+1) + (p+1)^(2k-1) = (p^(2) + p + 1) m "....."(1)`
Now, `p^(k+1) + (p + 1)^(2k+1)`
`= p xx p^(k+1) + (p + 1)^(2k-1) xx (p+1)^(2)`
`= p [m(p^(2) + p + 1 ) - (p + 1)^(2k-1)] + (p+1)^(2k-1)(p+1)^(2)`
`= p(p^(2) + p + 1) m - p (p+1)^(2k-1) + (p+1)^(2k-1)(p^(2) + 2p+1)`
`= p (p^(2) + p + 1) m + (p+1)^(2k-1)(p^(2) + p + 1)`
`= (p^(2) + p + 1 ) [mp+(p+1)^(2k-1)]`
`= (p^(2) + p + 1) xx ("an interger")`
Therefore, `P(k+1)` is also true whennever`P(k)` is true.
Hence, by principle of mathematical induction, `P(n)` is true `AA n in N`.