Find all $n \in \mathbb{Z}$ such that $ k = \frac{1+4n}{5}, \qquad (k \in \mathbb{Z} )$

Question

My question is rather general but I got stuck in that issue after trying to solve a trigonometric equation.

After simplifying I got this:

$$\sin \left(\frac{5x}{4}\right) + \cos x = 2$$

which is true if and only if

\begin{align*} \begin{cases} \sin \left(\frac{5x}{4}\right) &= 1\\ \cos x &= 1 \end{cases} \end{align*}

Hence

\begin{align*} \cos x &= 1 \\ \implies x &= 2\pi k, \qquad (k \in \mathbb{Z} ) \end{align*}

and

\begin{align*} \sin \left(\frac{5x}{4}\right) &= 1\\ \implies \frac{5x}{4} &= \frac{\pi}{2} + 2\pi n\\ &= \frac{2\pi}{5} + \frac{8\pi n}{5},\qquad (n \in \mathbb{Z} ) \end{align*}

Thus the solutions are

\begin{align*} 2\pi k &= \frac{2\pi}{5} + \frac{8\pi n}{5}\\ k &= \frac{1+4n}{5}, \qquad (k \in \mathbb{Z} ) \end{align*}

Here I'm stuck. Clearly the constraint of $k \in \mathbb{Z}$ is not true for all $n$'s. $n$ has to be made up of some number $m$ that makes the numerator divisible by $5$. The answer that I'm given says that $n = 5m + 1$, which makes sense but I don't know how to get there. And what if, just to say, $\displaystyle k = \frac{8+13n}{7}$?

Thanks!!

Answer 1

If you're not familiar with congruences, you can look at it this way: If $4n+1$ is divisible by $5$, then there is some integer $r$ with $4n+1 = 5r$, so that $4n = 5r-1$ and then $4n-4=5r-5$. Factoring both sides gives $4(n-1) = 5(r-1)$. Since $4$ and $5$ are relatively prime, it follows that $5$ divides $n-1$; that is, $n-1 = 5s$ for some integer $s$ and thus $n = 1+5s$.

Answer 2

$k$ is an integer if and only if $n=5m+1$ for some $m\in \mathbb Z$

To see this, solve the equation $$4n+1\equiv 0\ (\ mod\ 5\ )$$

This is equivalent to $$4n-4=4(n-1)\equiv 0\ (\ mod\ 5)$$ , which immediately implies that $n\equiv 1\ (\ mod\ 5\ )$ is necessary and sufficient.

Answer 3

$4n+1$ divisible by 5 $\Leftrightarrow 4n+1 \equiv_{5} 0 \Leftrightarrow -n \equiv_{5} -1 \Leftrightarrow n \equiv_{5} 1$ So n is an integer of the form $1+5q$ for abitrary integer $q$.