If $\frac{5m-1}{4} =n$ m,n $\in$ Z integer then find general solution [closed]

If $$\dfrac{5m-1}{4} =n$$ m,n $$\in$$ Z (integer) then find general solution. Answer is m = 4k -3, n= 5k-4. $$k\in$$ Z(integer) According to me answer should be 1+4n=5k , 5m-1=4k. Can you please explain where did I go wrong ? I am a high school students who is self studying for college entrance exam. Original problem was a trigonometric equation which I solved but didn't get the correct answer. Thanks you.

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• You wrote $1+4n=5k$ but should be $1+4n=5m$; also you wrote $5m-1=4k$ but should be $5m-1=4n$ – J. W. Tanner Apr 11 at 2:43
• @J.W.Tanner it is suppose to be in general form , m should be independent of n and vice versa. So I needed another variable integer. – user541396 Apr 11 at 2:46
• Why is this question being downvoted? – user541396 Apr 11 at 2:46
• math.stackexchange.com/questions/3182436/… – lab bhattacharjee Apr 11 at 3:34
• I would not say $m$ is independent of $n$ if $\dfrac{5m-1}{4} =n$ – J. W. Tanner Apr 11 at 3:46

If $$\dfrac{5m-1}{4} =n$$ with $$m,n \in \mathbb Z$$ , then $$m$$ and $$n$$ are not independent of each other;

they satisfy $$1+4n=5m$$ (you wrote $$5k$$) or $$5m-1=4n$$ (you wrote $$4k$$).

From $$1+4n=5m$$ we see $$5$$ divides $$4n+1$$ so $$5$$ divides $$4n+1+15=4n+16=4(n+4)$$

so $$5$$ divides $$n+4$$ (since $$5$$ is prime and does not divide $$4$$) so $$5k=n+4$$ so $$n=5k-4$$ with $$k\in\mathbb Z$$.

Therefore $$m=\dfrac{1+4n}5=\dfrac{1+4(5k-4)}5=\dfrac{20k-15}5 =4k-3.$$

• Alternatively, I could have said $5m-1=4n \implies 4n\equiv-1\pmod5\implies n\equiv1\pmod5 \implies n=5k-4$ so $m=\dfrac{1+4n}5 = \dfrac{1+20k-16}5 = \dfrac{20k-15}5=4k-3$ with $k\in \mathbb Z$. – J. W. Tanner Apr 11 at 3:41
• Can you please explain what "$5 m \equiv 1 (mod 4) "$ means. It's the first time I am seeing this kind of expression. I thought $\equiv and =$ mean basically the same thing. – user541396 Apr 11 at 4:10
• $5m\equiv1\pmod4$ means $4$ divides $5m-1$ – J. W. Tanner Apr 11 at 4:11
• $k = K+1$ looks arbitrary. Can you please explain that. – user541396 Apr 11 at 4:22
• $4K+1$ is the same as $4k-3$ -- I merely wanted to switch to the form you said was the answer -- but anyway I revamped my answer so I no longer use modulo notation or $K$ – J. W. Tanner Apr 11 at 4:25