Explanation:
I think you mean
p ≡ 1 mod 6 or p ≡ 5 mod 6
I will base my answer on this assumption.
remember what p ≡ n modulo m means :
after p/m there is a remainder n.
so, n can only be 0, 1, 2, 3, ..., m-1
in our case that means n can be 0, 1, 2, 3, 4, 5.
so, let's go through each case :
0 mod 6 would mean there is no remainder after the division. so, clearly p was divisible by 6 and is no prime number.
1 mod 6 : if after a division by 6 there is a remainder of 1, it means p is at least an uneven number, and every prime number greater or equal to 5 is uneven (odd).
because (p-1) is then 0 mod 6 and therefore an even number, (p-1) plus 1 (= p) must therefore be an uneven number.
2 mod 6 : p might not be divisible by 6, but it must be an even number and therefore cannot be a prime number.
why, you ask ? if p is uneven and has a remainder of 2, that means (p-2) must be divisible by 6. but an uneven number minus 2 is again an uneven number. and an uneven number cannot be divided by 6 (as 6 contains 2 as prime factor). so, p has to be even.
3 mod 6 : it means p is not divisible by 6, but it is divisible by 3 (as 3 is half of 6, and 6 = 2×3). if (p-3) is divisible by 6, it is also divisible by 3 (and by 2). adding 3 to it (giving us p) making it again divisible by 3.
4 mod 6 : similar to 2 mod 6. (p-4) is divisible by 6 making it an even number. adding 4 (giving us p) results again in an even number. therefore, not a prime number.
5 mod 6 : similar to 1 mod 6. we could say it is -1 mod 6. it means that (p-5) is divisible by 6 and therefore an even number. adding 5 to an even number makes it uneven. and therefore a possible prime number.
because we could rule out 0, 2, 3 and 4 as possible remainders after dividing a prime number by 6, and only 1 and 5 are possible remainders for a prime number divided by 6, we have proven the given statement.