Answer
When $n=1$, the value of $n^2-n+41$ is $41$, which is a prime number.
However when $n=41$, the value of $n^2-n+41$ is $41$ is $4162$, which is not prime.
Thus, the statement :$``n^2 - n + 41\text{ is a prime number"}$ is not true.
Work Step by Step
We need to prove that $p(n): n^2-n+41$ is a prime number.
Let us check for $n=1$ that $p(1)$ is true.
So, we have: $n^2-n+41=1-1+41=41$, this shows that $p(1)$ is a true and $41$ is prime.
Let us check for $n=41$ that $p(41)$ is true.
So, we have: $n^2-n+41=41^2-41+41=(41)^2$, this shows that $p(41)$ is not true and $(41)^2$ is not prime.
(because it has $41$ as a factor apart from $1$ and itself.
Thus, the statement $p(n): n^2-n+41$ is not true.
