When the following is written:
a ≡ b (mod p)
can it be written as
a ≡ b mod p
or
a ≡ b modulo p
as well? (all 3 forms are ok?)
I also saw
x ≡ 7 ≡ y (mod 11)
in this article, which started to make it confusing.
Is the (mod p) like a side note saying "hey, there is a world or universe in which we only consider the remainder after dividing by p, and in this world we say a is equal to b or a is congruent to b by the notation: a ≡ b or a = b", or is the mod some sort of operator like +, -, or in computer programming, the % operator so that 5 % 2 = 1 or 12 % 5 = 2?
P.S. Later on, I found that you can even say, 3 is equivalent to 1, modulo 2, for 3 ≡ 1 (mod 2). The term "equivalent" is used. (This is in Cormen's MIT Algorithms book, 3rd Edition, page 54). So I suppose 3 is equivalent to 1, in the world of modulo 2.
$a\equiv b\pmod p$ means "group all integers into equivalence classes depending on their remainder after division by $p$. Both $a$ and $b$ belong to the same equivalence class."
So you are right that the "mod $p$" in parentheses is a comment about the notation $\equiv$, and not an operator being applied to $b$. Incidentally I do recommend writing the parentheses, as $$a\equiv b \bmod p$$ has the potential of being confused with $$a=b\bmod p$$ which most people would interpret as meaning, "divide $b$ by $p$ and take the remainder. This remainder and $a$ are equal, as ordinary integers."