Question

Show that the statement \(" n^{2}-n+41\) is a prime number" is true for \(n=1\) but is not true for \(n=41\).

Short answer

True for \(n=1\). False for \(n=41\).

Step-by-step solution

Substitute n=1

Substitute the value of 1 for n in the expression \(n^{2}-n+41\). This gives \(1^{2} - 1 + 41\).

Simplify for n=1

Simplify the expression \(1^{2} - 1 + 41\) to get \(1 - 1 + 41 = 41\). Since 41 is a prime number, the statement is true for \(n=1\).

Substitute n=41

Substitute the value of 41 for n in the expression \(n^{2}-n+41\). This gives \(41^{2} - 41 + 41\).

Simplify for n=41

Simplify the expression \(41^{2} - 41 + 41\) to get \(41^{2} = 1681\). Since 1681 is not a prime number \((1681 = 41 \times 41)\), the statement is not true for \(n=41\).

Key concepts

Polynomial Evaluation

Polynomial evaluation is a method used to substitute a value for the variable in a polynomial and then simplify it. In our example, the polynomial is given by: \[ n^{2} - n + 41 \] We evaluated this polynomial at two points: \( n = 1 \) and \( n = 41 \).
When we substitute \( n = 1 \) into the polynomial, we get: \( 1^{2} - 1 + 41 \).
This simplifies to \( 41 \).
For \( n = 41 \), the polynomial becomes: \( 41^{2} - 41 + 41 \), which simplifies to \( 41^{2} = 1681 \).
Thus, polynomial evaluation helps us understand the behavior of the polynomial at specific values of \( n \).

Prime Number Verification

Prime number verification is the process of determining if a number is a prime. A prime number is a number greater than 1 that has no positive divisors other than 1 and itself. Let's apply this to our solutions:

- For \( n = 1 \), the result is 41. To verify if 41 is prime, we check divisibility by primes less than \( \sqrt{41} \) (approximately 6.4). Since 41 is not divisible by 2, 3, 5, 7, etc., it is indeed prime.

• For 41: It’s prime because no numbers other than 1 and 41 divide it.
  

- For \( n = 41 \), the result is 1681. To verify if 1681 is prime, we check smaller divisors. Since \( 41 \times 41 = 1681 \), it proves that 1681 has divisors other than 1 and itself and is thus not prime.

Algebraic Substitution

Algebraic substitution involves replacing the variable in the expression with a number. The process is straightforward:
1. Identify the polynomial or algebraic expression. Here it’s \( n^{2} - n + 41 \).
2. Choose the value to substitute (e.g., \( n = 1 \) or \( n = 41 \)).
3. Replace \( n \) in the polynomial with the chosen value.

• Substituting \( n = 1 \): \( 1^{2} - 1 + 41 \) results in 41.
  
• Substituting \( n = 41 \): \( 41^{2} - 41 + 41 \) simplifies to 1681.
  

Algebraic substitution helps in evaluating the polynomial at different values, unveiling insights about its behavior under various inputs.