Find all $x\in \mathbb{Z}$ satisfying the equation $5x+1 \equiv 13 ~~{\rm(mod~23)}$

Question

I am doing some exercises on groups and I am having trouble trying to show why the solution to: $$5x+1 \equiv 13 ~~{\rm(mod~26)}$$ is $$18 + 26\mathbb{Z}$$.

Can anyone please explain? Thanks for your help.

Answer 1

we have $$5x\equiv 12 \mod 26$$ and this is equivalent to $$x\equiv \frac{12}{5}\equiv \frac{48}{5}\equiv\frac{74}{5}\equiv \frac{100}{5}\equiv 20 \mod 26$$

Answer 2

Note that $5\cdot 5 \equiv -1\pmod{26} \implies 5^{-1} \pmod{26} \equiv -5\pmod{26}$. Hence, we have $$5x+1 \equiv 13\pmod{26} \implies 5x \equiv 12\pmod{26}$$ $$5x \equiv 12\pmod{26} \implies x \equiv-60\pmod{26} \equiv 18\pmod{26}$$

Answer 3

Simply done :

For some $y$

$$5x = 12 + 26y$$

So taking mod 5 we can infer that for some $n$ :

$$5n = 2 + y$$

$$y = 5n - 2$$

And substituting back we get :

$$5x = 12 + 5(26)n - 26(-2)$$

$$5x = -40 + 5(26)n$$

$$x = -8 + 26n$$

And we can rewrite that as by setting $n = m+1$

$$x = 18 + 26m$$