The statement suggests that $n^2 + n + 1$ is a prime number for all $n \geq 1$. To disprove this, we need to find a value of $n \geq 1$ such that $n^2 + n + 1$ is not a prime number.

Let's check a few values of $n$:

• For $n = 1$: $n^2 + n + 1 = 1^2 + 1 + 1 = 3 \quad (\text{prime})$

• For $n = 2$: $n^2 + n + 1 = 2^2 + 2 + 1 = 7 \quad (\text{prime})$

• For $n = 3$: $n^2 + n + 1 = 3^2 + 3 + 1 = 13 \quad (\text{prime})$

• For $n = 4$: $n^2 + n + 1 = 4^2 + 4 + 1 = 21 \quad (\text{not prime, since } 21 = 3 \times 7)$

Thus, when $n = 4$, the expression $n^2 + n + 1$ gives 21, which is not a prime number. Therefore, the value of $n = 4$ disproves the statement.

Would you like further details or have any questions?

Related Questions:

1. What is a prime number, and how do we identify it?
2. Can you prove that $n^2 + n + 1$ is prime for other small values of $n$?
3. What is the general method to find when quadratic expressions are prime?
4. Are there any other patterns in numbers that lead to prime results?
5. How can you factor $n^2 + n + 1$ for specific values of $n$?

Tip:

When testing mathematical statements, start with small values to check if a counterexample exists quickly.