Question

If $P\left( n \right)$ is the statement ${n^2} - n + 41$ is prime, prove that $P\left( 1 \right)$, $P\left( 2 \right)$ and $P\left( 3 \right)$ are true.
Prove also that $P\left( {41} \right)$ is not true.

Answer 1

Hint: To prove this, we will substitute different values of $n$ in the given statement of $P\left( n \right)$ . If the number that we will get after substituting the value of $n$ is a prime number, then the statement is true; otherwise, if the number is not a prime number, then the statement is false,

Complete step-by-step answer:
It is given in the question that $P\left( n \right) = {n^2} - n + 41$ is a prime number. We can say that according to this statement, when we have different values of $n$ in the given number, then for that value of $P\left( n \right)$ must be a prime number.
To find $P\left( 1 \right)$ , we will substitute 1 for $n$ in the expression ${n^2} - n + 41$ . This can be expressed as:
$\begin{array}{l}
P\left( 1 \right) = {1^2} - 1 + 41\\
P\left( 1 \right) = 41
\end{array}$
We know that 41 is a prime number; hence this statement is true for $P\left( 1 \right)$.
Now to find $P\left( 2 \right)$, we will substitute 2 for $n$ in the expression ${n^2} - n + 41$ . This can be expressed as:
$\begin{array}{l}
P\left( 2 \right) = {2^2} - 2 + 41\\
P\left( 2 \right) = 4 - 2 + 41\\
P\left( 2 \right) = 43
\end{array}$
We know that 43 is a prime number hence this statement is true for $P\left( 2 \right)$.
Now to find $P\left( 3 \right)$ , we will substitute 3 for $n$ in the expression ${n^2} - n + 41$ . This can be expressed as:
$\begin{array}{l}
P\left( 3 \right) = {3^2} - 3 + 41\\
P\left( 3 \right) = 6 - 3 + 41\\
P\left( 3 \right) = 47
\end{array}$
We know that 47 is a prime number hence this statement is true for $P\left( 3 \right)$.
Now to find $P\left( {41} \right)$ , we will substitute 41 for $n$ in the expression ${n^2} - n + 41$ . This can be expressed as:
$\begin{array}{l}
P\left( {41} \right) = {41^2} - 41 + 41\\
P\left( {41} \right) = {41^2}\\
P\left( {41} \right) = 1681
\end{array}$
Since 1681 is the square of 41, hence it is not a prime number. Hence we can say that this statement is not true for $P\left( {41} \right)$.

Note: Prime numbers can be considered as the numbers which are only divisible by 1 or by itself. Hence in this question, we will check for the numbers which are only divisible by 1 and itself. To check whether the number is prime or not, we can perform the divisibility test of 2, 3, 5, 7 and 11 on the numbers.