2. Theorem
The only solutions in integers is a prime of the equation are given by the diophantine equation , p=3, n=5, p=5, n=7 so n is integer and p is prime number.
Ist Proof
Given the Diophantine equation , where p and n are prime numbers.
We need to show that the only solutions in positive integers for this equation are when and , or when and .
Let’s assume there exists another solution where and are prime numbers, and .
First, observe that for any prime , since cycles through modulo 3. Thus, , which implies , and . However, this implies , as 3 is the only prime equivalent to 0 modulo 3. Thus, any solution must have .
Now, let’s consider the parity of n. If n is even, then is a perfect square, and would also have to be a perfect square. However, for any prime , , and since no perfect squares are congruent to 3 modulo 4, cannot be a perfect square. Therefore, n must be odd.
When n is odd, is not a perfect square, which means cannot be a perfect square either. This implies that , as any odd perfect square is congruent to 1 modulo 4. This further implies , which is not possible since perfect squares are congruent to either 0 or 1 modulo 4. Hence, there are no solutions where n is odd.
Therefore, the only solutions are when or . By checking these values, we find that and , so or . Thus, the only solutions in positive integers are and , as claimed. Hence, the proof is complete.
2nd Proof
To prove that the only solutions in integers for the equation are and , and where p is a prime number and n is an integer, we will employ a proof by contradiction.
Suppose there exists another solution where p is a prime number and n is an integer such that , and .
First, note that for any integer n, if n is even and if n is odd. This is because , and therefore .
Considering , we have:
If , then , and . If , then , and . If , then , and . Since or for any integer n, there cannot exist a prime p such that . Therefore, cannot be divisible by 3.
Now, let’s examine modulo 4. Note that for any integer p, if p is even, and if p is odd.
If p is even, then , and . If p is odd, then , and .
Since for any integer n, there cannot exist a prime p such that . Therefore, cannot be congruent to 3 modulo 4.
From the above, we conclude that cannot be divisible by 3 and cannot be congruent to 3 modulo 4. Thus, there are no solutions other than and .
This completes the proof by contradiction. Therefore, the only solutions in integers for the equation are indeed and , and .