Question

Find all the prime numbers of the form ${{n}^{3}}-1$, where n is a natural number.

Answer 1

Hint: To solve this question, we will see what is the meaning of Mersenne prime numbers. And, we will see what are the first four Mersenne Prime numbers, and using the standard form of Mersenne Prime number, we will find prime numbers of form ${{n}^{3}}-1$.

Complete step-by-step solution:
Now, in mathematics, a Mersenne Prime is a prime number that is on less than a power of two . that is prime number of the form ${{2}^{n}}-1$, for some integer n. there are total 51 mersenne prime number known.
First four terms of Mersenne Prime of form ${{n}^{3}}-1$ are 3, 7, 31 and 127.
Now, in question it is asked to find all the prime number of form ${{n}^{3}}-1$.
Now, comparing ${{n}^{3}}-1$ with mersenne prime, we get integral power equals to 3.
So, putting, n = 2 in mersenne prime ${{n}^{3}}-1$, we get
${{2}^{3}}-1=8-1=7$, we get 7 which is a prime number as 7 is only divisible by 1 and itself.
So, total number of prime numbers of form ${{n}^{3}}-1$, is only one prime which is7.
But, if take any another integer say, n = 4
So, ${{4}^{3}}-1=64-1=63$, which is not prime as 63 is divisible by 1, 3, 7, 9 and 63 and prime numbers are only divisible by 1 and itself only.
Hence, 7 is only a prime number of the form ${{n}^{3}}-1$.

Note: To find the prime number of form ${{n}^{3}}-1$, one must know the special Mersenne Prime known as ${{2}^{n}}-1$. This question can be solved by factorising , by using formula ${{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})$ , then solve by putting both two factors equals to 1, which will give only one solution n = 2, and putting n = 2 in ${{n}^{3}}-1$, we will get only one prime which is equals to 7.