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Hi I am studying modular arithmetic just for myself, but to be honest I find it very difficult. I am not sure which answer is right. Is it right to say

If $4x \equiv 8\pmod{15}$ then $x \equiv 2 \pmod{15}$ or just $x=2$ ?

Thank you

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  • $\begingroup$ use \equiv for $\equiv$ $\endgroup$ – ganeshie8 Oct 2 '14 at 12:16
  • $\begingroup$ Apologies there should be an 8 on the RHS $\endgroup$ – moony Oct 2 '14 at 12:22
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Hint: See what happens when $x = 17$, for example.

The reason you can deduce $x\equiv 2\pmod{15}$ from $4x \equiv 8\pmod{15}$ is that $4\times 4 \equiv 1 \pmod{15}$. If $4x\equiv 8 \pmod{15}$ then also $16x\equiv 32\pmod{15}$ (multiplying by $4$: this is legal since if $15|4x-8$ then also $15|4(4x-8)$). However, $16\equiv 1\pmod{15}$ and $32\equiv 2\pmod{15}$, this implies $x\equiv 2\pmod{15}$ (this is a bit harder to justify, but hopefully you saw this in class).

We can say a little more: if $x \equiv 2\pmod{15}$ then multiplying by $4$, we get $4x\equiv 8\pmod{15}$. So both statements are equivalent. This means that if you take any $x$ satisfying $x\equiv 2\pmod{15}$ then you will have $4x \equiv 8\pmod{15}$.

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The question is not well-formed: in modular arithmetic, $$ a \equiv b \mod(n) \iff n \text{ divides } a - b $$

In your case, you are not using any number/entity as "$b$" in the first equation, so the equations you wrote are incomplete, so the question does not have sense.

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  • $\begingroup$ Yes, apologies,there should be an 8 there $\endgroup$ – moony Oct 2 '14 at 12:23
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$$4x\equiv 8\pmod{15}\iff x\equiv 2\pmod{15}.$$ This is true because $4$ (by which you divide the both sides) and $15$ are coprime. (otherwise you cannot divide the both sides.) But note that $x=2$ is not the only solution. The answer is $x=15k+2$ where $k\in\mathbb Z$. For example, $x=17,32,47,\cdots$ satisfy $x\equiv 2\pmod{15}$.

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  • $\begingroup$ Thank you all so very very much $\endgroup$ – moony Oct 2 '14 at 12:39
  • $\begingroup$ @moony: You are welcome. $\endgroup$ – mathlove Oct 2 '14 at 12:40
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Hopefully it helps to make a table

n | 4n mod 15

0 | 0

1 | 4

2 | 8

3 | 12

4 | 1

5 | 5

6 | 9

7 | 13

8 | 2

9 | 6

10| 10

11| 14

12| 3

13| 7

14| 11

So there is a one to one correspondence between n in {0..14} and 4*n mod 15.

x is congruent to 2 mod 15 is different from x = 15 because the first one includes 17.

Regards, Matt

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  • $\begingroup$ ok thank you, I'm still unsure on which is the right answer though ? $\endgroup$ – moony Oct 2 '14 at 12:29
  • $\begingroup$ x is congruent to 2 mod 15 is a more complete answer. For example, if I ask "What x gives solution to x^2 + 7*x + 10 = 0?" The answer x = -2 is correct but x = -2 or -5 is more complete. $\endgroup$ – user145600 Oct 2 '14 at 13:04

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