Question

Find all primes p such that 17p+1 is the square of an integer. Hint: If 17p+1= na then 17p = (n − 1)(n+1).

Answer 1

To find all primes p such that 17p+1 is the square of an integer, use the hint provided. Solve the equation 17p = (n-1)(n+1) to find the prime values of p that satisfy this equation. The prime values of p that satisfy the equation are p = 1 and p = 51.

Step-by-step explanation:

To find all primes p such that 17p+1 is the square of an integer, we can use the hint provided. If 17p+1 = na where n is an integer, then 17p = (n-1)(n+1). Now we need to find the prime values of p that satisfy this equation.

We know that 17p is equal to the product of (n-1) and (n+1). Since p is prime, it can only be divided evenly by 1 and itself. Therefore, (n-1) and (n+1) must also be factors of 17p.

Since p is prime, it cannot be divided evenly by any number other than 1 and itself. Therefore, (n-1) and (n+1) must be factors of 17, and 17 can only be factored as 1 and 17.

From this, we can deduce that (n-1) and (n+1) must be 1 and 17, or vice versa. Solving these equations, we find that n = 2 and n = 18.

Substituting these values back into the equation 17p = (n-1)(n+1), we find that when n = 2, p = 1, and when n = 18, p = 51.

Therefore, the only prime values of p that satisfy the given equation are p = 1 and p = 51.

Answer 2

No prime numbers p exist such that 17p+1 is the square of an integer because for any prime number p, there are no consecutive even numbers that can be multiplied to equal 17p.

Step-by-step explanation:

To find all primes p such that 17p+1 is the square of an integer, we follow the hint provided. If 17p+1 = n², then 17p = (n - 1)(n + 1). Since n - 1 and n + 1 are two consecutive even numbers, the only way they can be factors of 17 (a prime number) is if one of them is a multiple of 17 and the other is 1 (since 17 × 1 is the only way to express 17 as a product of natural numbers due to its primality). However, the difference between any two consecutive even numbers is 2, not 17, so there are no such consecutive even numbers n - 1 and n + 1 which could satisfy this condition. Therefore, there are no prime numbers p for which 17p+1 is the square of an integer.