Problem:
Prove that p^2 – 1 is divisible by 24 if p is a prime number greater than 3?
Solution:
We can factorize p^2 - 1 to (p-1)*(p+1) = 2*3.4(k), k is some integer
The most elementary proof , without explicitly mentioning any number theory: out of the three consecutive numbers p–1, p, p+1, one of them must be divisible by 3; also, since the neighbors of p are consecutive even numbers, one of them must be divisible by 2 and the other by 4, so their product is divisible by 3⋅2⋅4=24 — and of course, we can throw p out since it’s prime, and those factors cannot come from it.
Prove that p^2 – 1 is divisible by 24 if p is a prime number greater than 3?
Solution:
We can factorize p^2 - 1 to (p-1)*(p+1) = 2*3.4(k), k is some integer
The most elementary proof , without explicitly mentioning any number theory: out of the three consecutive numbers p–1, p, p+1, one of them must be divisible by 3; also, since the neighbors of p are consecutive even numbers, one of them must be divisible by 2 and the other by 4, so their product is divisible by 3⋅2⋅4=24 — and of course, we can throw p out since it’s prime, and those factors cannot come from it.
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