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Can't we do something like this:

$2n = 6k - 1$ where k is an integer and n is a normal number

We treat $k' = 2k$ which means

$2n = 3k' - 1$

And by using modulo arithmetic

$2n = -1 (mod 3)$

Which is $n = 1 (mod 3)$ since 2 and 3 are co-prime.

Then the solution is that $n = 3p + 1$ where p is an integer.

The book says that it has no solution but isn't this a solution?

I'm definitely wrong but I applied this solution to many problems before, I dunno why I can't apply it here?

EDIT: I'm so sorry, I mean $n$ is a natural number not normal.

Bill Dubuque
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TechnoKnight
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    It has a solution modulo $3$, but no solution over the integers. The solution that comes out modulo $3$ can't be interpreted as an integer solution. – Sarvesh Ravichandran Iyer Jan 27 '21 at 15:23
  • @TeresaLisbon ? – TechnoKnight Jan 27 '21 at 15:24
  • @TeresaLisbon why it can't be a solution? – TechnoKnight Jan 27 '21 at 15:25
  • By normal number, do you mean this? – Prasun Biswas Jan 27 '21 at 15:25
  • @PrasunBiswas Yes. – TechnoKnight Jan 27 '21 at 15:26
  • @TechnoKnight It cannot be a solution, because $2n - 1$ is a multiple of $3$, that's all you know. You don't know that it is a multiple of $6$, which is what $2n -1 = 6k$ means. By setting $k' = 2k$ you have forgotten that $k'$ itself must be even. – Sarvesh Ravichandran Iyer Jan 27 '21 at 15:26
  • @PrasunBiswas Wait, no. I mean a number without , and just positive. I think it's called natural number in English? – TechnoKnight Jan 27 '21 at 15:27
  • Then my deleted comment was correct: LHS is even but RHS is odd, so no solutions over the integers. – Prasun Biswas Jan 27 '21 at 15:28
  • Ok, let me say it again : we must find a solution to $2n+1 = 6k$ for $n,k$ integers. Now, you took $2k = k'$ so now you want to solve $2n+1 = 3k'$ for $n$ integer and $k'$ an EVEN integer. Writing this as $2n = - 1 \pmod{3}$ , solves the equation $2n-1 = 3l$ for all $l$, because you are only concerned about whether $2n-1$ is a multiple of $3$ or not, while in truth you should be concerned about whether it is an even multiple or not. This reduction gives you solutions that aren't true to the original problem, because in each of these, $k'$ is odd, while you want it to be even. – Sarvesh Ravichandran Iyer Jan 27 '21 at 15:30
  • Getting a solution modulo $n$ doesn't imply solution over the integers: consider $2x=1$; this has a solution modulo $3$ (any integer with residue 2 mod 3) but over integers, we have $2x=1\implies x=1/2\notin\Bbb Z$ – Prasun Biswas Jan 27 '21 at 15:31
  • @PrasunBiswas I might be a retard(I'm sorry) but I still don't get it. When I know that what can be a solution and what's not? I mean, I applied this way to a lot of stuff and they were all correct. Take 11x−24y=1 as example. You use Modulo arithmetic and you find x=24k+11 and y=11k+5 where k is an integer. Why I can use it here and not on that solution? – TechnoKnight Jan 27 '21 at 15:40
  • @Vepir I do understand that my solution is completely wrong. But what I don't understand is why we use Modulo arithmetic to solve equations like $11x - 24y = 1$ but not that equation in my question? – TechnoKnight Jan 27 '21 at 15:57
  • @TechnoKnight You can use it to solve your question, if you use it correctly. $n=3p+1$ is not a solution, it is a requirement for a solution. Another requirement is 0=1 (if you observe the equation modulo 2), which is clearly always false regardless of n. Since it is impossible to satisfy all requirements, there are no solutions. – Vepir Jan 27 '21 at 16:41

4 Answers4

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The left hand side of your equation is even. The right hand side is odd. It's going to be very tough to find a number that's both even and odd.

Duncan Ramage
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  • But didn't I find a solution at the end? $n= 3p +1$ isn't a solution? – TechnoKnight Jan 27 '21 at 15:26
  • I wrote that in a comment earlier but deleted after seeing "$n$ is a normal number" – Prasun Biswas Jan 27 '21 at 15:26
  • @TechnoKnight No, you didn't. You found another expression that a solution would have to satisfy. – Duncan Ramage Jan 27 '21 at 15:27
  • @DuncanRamage But then it's easily satisfied? I mean 7 works(Since 7 = 3(2) + 1) – TechnoKnight Jan 27 '21 at 15:29
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    @TechnoKnight But 7 isn't a solution to your equation. You have the logical implication backwards. (n is a solution) implies (n = 3p + 1). Just because you can find numbers that are 1 mod 3 doesn't mean they are, in fact, solutions. – Duncan Ramage Jan 27 '21 at 15:31
  • @DuncanRamage I might be a retard(I'm sorry) but I still don't get it. When I know that what can be a solution and what's not? I mean, I applied this way to a lot of stuff and they were all correct. Take $11x - 24y = 1$ as example. You use Modulo arithmetic and you find $x = 24k + 11$ and $y = 11k + 5$ where k is an integer. Why I can use it here and not on that solution? – TechnoKnight Jan 27 '21 at 15:34
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$\begin{align} {\bf Hint}\quad\ 6\mid 2n\!+\!1 &\iff\ \ \ \ 3\mid 2n\!+\!1\ \text{ is false (your assumption)}\\[.2em] 6\mid 2n\!+\!1 &\iff \color{#c00}2,3\mid 2n\!+\!1\ \text{ is true (by lcm or CRT)} \end{align}$

Renark $ $ It is generally true that congruences persist mod factors of the modulus (i.e. that the direction $(\Rightarrow)$ above holds true), and this arrow doesn't reverse for proper factors of the modulus (because it doesn't capture the divisibility information implied by the deleted factors of the modulus, e.g. deleting the factor $\,\color{#c00}2\,$ above loses parity information, which is crucial there).

Bill Dubuque
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  • I think I understand now. You mean that by doing my way, I'm assuming that only 3 should divide $2n - 1$ which is false because dividing by 3 doesn't mean that 6 divides it. So I should also test if 2 divides it and since 2 clearly can't because 2n - 1 is odd then the solution is clearly wrong and the equation has no solution. Is that what you're saying? – TechnoKnight Jan 27 '21 at 16:31
  • @TechnoKnight Yes, you correectly infer direction $(\Rightarrow)$ that $\ 6\mid 2n-1\Rightarrow 3\mid 2n -1.,$ so a solution of $,2n-1 = 6k,$ yields a solution of $2n-1 = 3k', $ (via $k' = 2k).,$ But this inference doesn't reverse to yield an equivalence $(!\iff!)$ because it ignores the factor of $,\color{#c00}2,$ (parity), which is where it fails. – Bill Dubuque Jan 27 '21 at 16:44
  • Okay, I understood that. But how do I know if can variable substitute or not if the equation is unknown to accept divisions? – TechnoKnight Jan 27 '21 at 16:45
  • @TechnoKnight To get an equivalent diophantine equation we need to ensure that the inverse of our change of variables yields integer solutions for the original variables. But yours does not: $,k = k'/2,$ need not be an integer. Otoh changes like variable shifts $,k' = k+n,$ do have integral inverses $,k = k' - n,$ so they always yield an equivalent diophantine equations. – Bill Dubuque Jan 27 '21 at 16:57
  • Okay I think I understand you better but how did you know that k = k'/2 is not an integer? – TechnoKnight Jan 27 '21 at 17:18
  • @TechnoKnight $,\color{#c00}{k'}+2k' = \color{#c00}1+2n\Rightarrow \color{#c00}{k' \equiv 1}\pmod{2} ,$ i.e. $,k',$ is odd. Generally we can't assume without justification that $,k'/2$ need be an integer. Indeed, it isn't in this case. – Bill Dubuque Jan 27 '21 at 17:29
  • Okay, I think I understand now. So to summary it up and see if I'm right, to variable substitute in a Diophantine equation I need to make sure that the variable I'm substituting is divisible by the equation(Like 10k, I need to make sure that both 5 and 2 divides the equation correctly so that I can change it to 5k' by example) and then see if the reverse gives us an integer or not. If I don't find one of these conditions correct then my solution is wrong and the equation has no solution. Is that correct? – TechnoKnight Jan 27 '21 at 17:39
  • @TechnoKnight In equation form: $,\exists, x!:\ 10:! x = n \iff \exists, x_1,x_2!:\ 5:!x_1 = n = 2:! x_2,,$ by LCM or CRT. But this is better viewed not in terms of the quotients $,x_i,$ but rather in terms of the equivalent divisibilities $,5\cdot 2\mid n \iff 5,2\mid n,,$ or, more arithmetically in operation (vs. relation) form using congruences $, n\equiv 0\pmod{!10}\iff n\equiv 0\pmod{!5\ &\ 2}.,$ – Bill Dubuque Jan 27 '21 at 18:35
  • One of the keys to solving such diophantine equations is learning how to quickly go back-and-forth between these forms, using whichever is most efficient in any context, e.g. see here. – Bill Dubuque Jan 27 '21 at 18:35
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So if $n=3p+1$ then $$6p+2=6k-1$$ or $$6(p-k)=3$$

Now both sides of this are divisible by $3$, so this is a valid equation modulo $3$, but $3$ is not a multiple of $6$.

Working modulo $3$ only captures part of the information in the original equation, so doesn't guarantee to find a solution of the original.

There is an interesting development in more advanced mathematics when one does look to solve equations modulo $p$ for primes $p$ (called working locally). Then one looks for conditions under which the existence of a local solution for every prime $p$ guarantees a global solution - ie a finite integer solution to the original. Note that even finding a solution for every prime does not guarantee a global solution (to more complex kinds of problems).

Mark Bennet
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You have:

$$2n+1=6k$$ so:

$$2n+1=2(3k)$$ if you put $3k=h$ you have:

$$2n+1=2h$$ So an odd number shoud be equal to an odd number which is impossible

Riccardo.Alestra
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