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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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closed as off-topic by Eevee Trainer, Lord Shark the Unknown, Leucippus, Adrian Keister, GNUSupporter 8964民主女神 地下教會 Apr 11 at 15:04

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  • $\begingroup$ You wrote $1+4n=5k$ but should be $1+4n=5m$; also you wrote $5m-1=4k$ but should be $5m-1=4n$ $\endgroup$ – J. W. Tanner Apr 11 at 2:43
  • $\begingroup$ @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. $\endgroup$ – user541396 Apr 11 at 2:46
  • $\begingroup$ Why is this question being downvoted? $\endgroup$ – user541396 Apr 11 at 2:46
  • $\begingroup$ math.stackexchange.com/questions/3182436/… $\endgroup$ – lab bhattacharjee Apr 11 at 3:34
  • $\begingroup$ I would not say $m$ is independent of $n$ if $\dfrac{5m-1}{4} =n$ $\endgroup$ – J. W. Tanner Apr 11 at 3:46
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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 $m=\dfrac{1+4n}5=\dfrac{1+4(5k-4)}5=\dfrac{20k-15}5 =4k-3.$

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  • $\begingroup$ 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 $k\in \mathbb Z$. $\endgroup$ – J. W. Tanner Apr 11 at 3:41
  • $\begingroup$ 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. $\endgroup$ – user541396 Apr 11 at 4:10
  • $\begingroup$ $5m\equiv1\pmod4$ means $4$ divides $5m-1$ $\endgroup$ – J. W. Tanner Apr 11 at 4:11
  • $\begingroup$ $k = K+1$ looks arbitrary. Can you please explain that. $\endgroup$ – user541396 Apr 11 at 4:22
  • $\begingroup$ $4K+1$ 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 $K$ $\endgroup$ – J. W. Tanner Apr 11 at 4:25

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