Modular arithmetic, very simple implications.

Question

$$3t \equiv 1 \mod 4 \Rightarrow t \equiv 3 \mod 4 $$

I don't understand that, so I'm asking for explain me.

Thank, in advance, greetings.

Answer 1

You know that $3$ and $4$ are coprime, so in particular, $3$ is invertible mod $4$.

What is the (unique) inverse? Well, there are only four candidates; and a quick check shows that

$$3 \times 0 \equiv 0 \mod 4,$$ $$3 \times 1 \equiv 3 \mod 4,$$ $$3 \times 2 \equiv 2 \mod 4,$$ $$3 \times 3 \equiv 1 \mod 4.$$

So the inverse of 3 modulo 4 is 3 itself. (Of course, we only really needed to check 1 and 3, as 0 and 2 are not invertible mod 4 and hence can't be the inverse of 3).

Thus when we multiply both sides of your first equation by 3, we get the answer.

Answer 2

$3t \equiv 1 \mod 4$ is simply means $3t-1$ is divisible by $4.$ $$3t \equiv 1 \mod 4 $$ $$3t \equiv (1+2\times 4) \mod 4$$ $$3t \equiv 9 \mod 4$$ and $(3,4)=1.$ Can you continue from here?

Answer 3

Saying $x\equiv y\pmod 4$ means that we can write $x$ as $y$ plus a multiple of $4$.

So we are assuming there is an integer $k$ with $3t=1+4k$. What we want to do then is to write $t$ as $3$ plus a multiple of $4$.

The trick is to multiply $3t=1+4k$ by $3$. This gives us $9t=3+12k$, so that $4(2t)+t=3+4(3k)$ and so $t=3+4(3k-2t)$. Since we have expressed $t$ as $3$ plus a multiple of $4$, we can say that $t\equiv 3\pmod 4$.

Answer 4

Another approach.

We know that $4t = 0 \pmod 4$. Then subtracting... $$3t = 1 \pmod 4 \ \Longrightarrow\ -t = 1 \pmod 4 \ \Longrightarrow\ t=-1=3\pmod 4$$