square ($a \bmod b$)

Question

suppose $a = np +r$ where $n$ is an integer, $p$ is a prime, $r$ is the remainder. So, $ a \bmod b ≡ r$.

Is $(a \bmod b)^2 = a^2 \bmod b^2$?

I was told this is correct, and can't prove it. (I don't want $a^2 \bmod b$, I am aware they mean different things).

$(a \bmod b)^2 $

$≡ (np +r)^2 $

$= (np)^2 + 2npr + r^2 $

$≡ 2npr + r^2 \bmod p^2 $

$\neq r^2 \bmod p^2 .$

Side note: Usually in number theory, modulo isn't an operation, but a relation. In abstract algebra, it's an operation that doesn't return a number, but a set of numbers: $a\bmod p={\ldots,a-p, a, a+p, a+2p,\ldots}$ — Arthur, Sep 06 '19 at 16:10
@Arthur: in this case, the notation should be $a\bmod p$ (\bmod). — Bernard, Sep 06 '19 at 16:13
If you mean $(a \bmod p)^2 = a^2 \bmod p^2$ instead of $(a \bmod b)^2 = a^2 \bmod b^2$, then this is not true. Take $a=5,p=3$. — mathlove, Sep 16 '19 at 04:58

score 2 · Accepted Answer · answered Sep 16 '19 at 06:38

This is not correct. Take $7=2\cdot 3 + 1$, where $a=7$, $n=2$, $p=3$, and $r=1$. Let $\text{%}$ denote your $\text{mod}$ operation. We have $(7\text{%} 3)^2=1^2=1$, but $7^2\text{%}3^2=49\text{%}9=4\neq 1$.

Of course, this equivalence is possible; take $10=3\cdot 3 + 1$. The equivalence happens when $2nr\equiv 0 \pmod p$. In this case, $2nr=6$, which is indeed divisible by $3$. However, this equivalence does not always hold true, as shown above.

square ($a \bmod b$)

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