Answer:
Explanation:
To prove that either p+1 or p-1 must always be divisible by 6 when p is a prime number greater than 3, we can use proof by contradiction.
Assume that both p+1 and p-1 are not divisible by 6. Then, by the division algorithm, we can write:
p = 6n + r
where n is an integer and r is the remainder when p is divided by 6. Since p is not divisible by 2 or 3, we know that r must be either 1 or 5.
If r = 1, then we have:
p - 1 = 6n
which means that p-1 is divisible by 6, contradicting our assumption.
If r = 5, then we have:
p + 1 = 6n + 6
which means that p+1 is divisible by 6, again contradicting our assumption.
Therefore, our assumption that p+1 and p-1 are not divisible by 6 must be false. Thus, either p+1 or p-1 must always be divisible by 6. ∎
To further explain why either p+1 or p-1 must always be divisible by 6, let's consider two cases:
Case 1: p is an even prime greater than 2. In this case, p+1 is odd and p-1 is even. Since p is even, it can be written as 2k for some integer k. Then we have:
p+1 = 2k+1 = 2(k+1/2)
Since k is an integer, k+1/2 is also an integer. Therefore, p+1 is divisible by 2.
p-1 = 2k-1 = 2(k-1/2) + 1
Since k is an integer, k-1/2 is also an integer. Therefore, p-1 is divisible by 2.
Now we need to prove that either p+1 or p-1 is divisible by 3. If p is not divisible by 3, then it can be written in the form of 3k+1 or 3k+2, where k is an integer.
If p = 3k+1, then p+1 = 3k+2 = 2(3k+1)/2, which is divisible by 2. Also, p+1 = 3k+2 is divisible by 3, since 3k+2 = 3(k+1)-1.
If p = 3k+2, then p-1 = 3k+1 is divisible by 3, since 3k+1 = 3(k+1/3)-2. Also, p-1 = 3k+1 is divisible by 2, since k is odd in this case.
Therefore, in either case, either p+1 or p-1 is divisible by 6.
Case 2: p is an odd prime. In this case, p+1 is even and p-1 is even. One of these numbers must be divisible by 2. We need to prove that one of them is also divisible by 3.
If p is not divisible by 3, then it can be written in the form of 3k+1 or 3k+2, where k is an integer.
If p = 3k+1, then p+1 = 3k+2 is divisible by 3, since 3k+2 = 3(k+1)-1. Also, p+1 = 3k+2 is divisible by 2, since k is odd in this case.
If p = 3k+2, then p-1 = 3k+1 is divisible by 3, since 3k+1 = 3(k+1/3)-2. Also, p-1 = 3k+1 is divisible by 2, since k is even in this case.
Therefore, in either case, either p+1 or p-1 is divisible by 6.
In conclusion, we have shown that for any prime number p greater than 3, either p+1 or p-1 is divisible by 6.