Is this correct $c\equiv a\pmod{b}$ $\implies \frac{c}2\equiv \frac{a}{2}\pmod{b}$

Question

I'm trying to understand if this way is correct always or not :

$a,b,c$ natural numbers

$b$ prime number

$a,c$ even number

if $\gcd (a,b)=1$

and we have

$c\equiv a\pmod{b}$

$\implies $ $\frac{c}{2}\equiv \frac{a}{2}\pmod{b}$

How I can understand and prove it ? Without using the Euclidean algorithm, because, I don't study $\frac{\mathbb{Z}}{p\mathbb{Z}}$

I know simple relations of modulo :

$c=b.k+c$ then $\frac{c}{2}=b\frac{k}{2}+\frac{a}{2} %only if a=c$

But may be $\frac{k}{2}\not\in\mathbb{N}$

My goal is to solve this :

$8n\equiv 1154\pmod{1163}$

So if we divide by $2$ we get

$4n\equiv 577\pmod{1163}$

$$4n\equiv 1740\pmod{1163}$$ Then

$$2n\equiv 870\pmod{1163}$$

Then

$$n\equiv 435\pmod{1163}$$ ( correct answer )

For example :

$20\equiv 1\pmod{19}$

$20\equiv 20\pmod{19}$

Then $10\equiv 10\pmod{19}$

Correct

So what's the proof ?

Answer 1

You can write $a=2a'$ and $c=2c'$ for some integers $a'$ and $c'$. Then $b$ divides $a-c=2(a'-b')$, so it must divide $2$ or divide $a'-c'$. It can't divide $2$ by assumption, then it divides $a'-c'$.

Answer 2

Let’s look at the more general statement:

If $a$, $b$ are even, and $a\equiv b\pmod n$, then $\frac a2\equiv\frac b2\pmod n$.

In general, this statement is false. $4\equiv2\pmod2$, but $2\not\equiv1\pmod2$.

Here’s the appropriate statement. If $a$, $b$ are even, and $n$ is odd, such that $a\equiv b\pmod n$, then it’s easy to prove that $$\frac a2\equiv\frac b2\pmod n,$$ as you suggest. However, if $n$ is even, the most you can say is $$\frac a2\equiv\frac b2\pmod{\frac n2}.$$ Translating the modulo condition into a divisibility condition and using some of its basic properties will be enough to prove both of these statements, and in particular, your question.

Answer 3

You're studying how to divide $\,a\,$ by $\,2,\,$ modulo odd $\,b,\,$ as in this answer to your prior question.

The idea is simple: either $\,a\,$ or $\,a\!-\!b\,$ is even. Call the even one $\,a' = 2k.\,$

This means $\!\bmod b\, $ we can always divide $\,a\,$ by $\,2\,$ by dividing $\,a'\,$ by $\,2,\,$ namely: if $\!\bmod b\!:\,\ 2x\equiv a\equiv a' \equiv 2k\iff 2x-2k = bn,\,$ so $\,bn\,$ even $\Rightarrow\, n\,$ even, by $\,b\,$ odd,

so, cancelling $\,2\,$ from above: $\iff x - k = b(n/2)\iff x\equiv k \pmod{\!b}$

In summary: $\ \ \dfrac{2k}2\equiv k\pmod{\!b}\ $ for odd $\,b$

and generally $\ \dfrac{ck}c\equiv k\pmod{b}\ $ when $\,\gcd(c,b)= 1,\ $ provable same as above,

i.e. $\,\ \bmod b\!:\ \ cx\equiv ck\iff x\equiv k,\,$ i.e. the fraction $\,x\equiv (ck)/k\,$ uniquely exists.

Indeed by Bezout $\,\gcd(c,b)=1\,\Rightarrow\,c^{-1}\,$ exists $\!\bmod b\,$ so we can cancel $\,c\,$ above by scaling by $\,c^{-1}\ $ (i.e. invertible elements are always cancellable, which means scaling equations and congruences by invertible elements always yields an equivalent one - similar to scaling a polynomial equation by the inverse of its leading coefficient to make it be $ 1$).

Remark $ $ More generally, as explained in the linked answer, we can compute a modular fraction (necessarily having its denominator coprime in order to the modulus to uniquely exist) by way of making the quotient exact, by adding or subtracting a suitable multiple of the modulus.

See this answer for general (multi-valued) modular division (and its fractional equivalent).