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4x mod 5 = 3x-1 mod 5

Should I do:




according to the theorem, it should be: if a mod 5 = b mod 5

then 5|(a-b), but shouldn't we take the greatest number?

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Hint $\rm\,\ m\mid k\iff m\mid -k\quad $ –  Bill Dubuque Dec 9 '12 at 0:03

2 Answers 2

It makes no difference at all. Suppose that $m\mid a-b$; then there is an integer $k$ such that $a-b=km$. But then $b-a=-(a-b)=-km$, and $-k$ is an integer, so $m\mid b-a$. Thus, the following four statements are all equivalent:

  1. $a\equiv b\pmod m$.
  2. $b\equiv a\pmod m$.
  3. $m\mid a-b$.
  4. $m\mid b-a$.
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This is good to know:

$$a\equiv b\pmod m \;\iff \;m\mid a-b \;\iff \; m\mid (b-a) \;\iff \; b\equiv a\pmod m \quad\quad\quad\tag{1}$$

$$ \text{and }\;\; m\mid c \quad\iff \quad m\mid -c \tag{2}$$

So what you've written is true, both statements:

$$5\mid (4x-(3x-1))\quad \text{AND}\quad 5\mid((3x-1)-4x)$$ That is: $$5\mid (x+1)\quad \iff \quad 5\mid -(x + 1)$$

So $x \equiv -1 \pmod 5\quad\iff \quad x\equiv 4 \pmod 5$.

Indeed, any $x = 5k + 4$ where $k$ is any integer is a solution: $x \in \{ \cdots, -6, -1, 4, 9, ...\}$.

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i get 5 = x-1 for the first and 5 = -x-1 for the second –  CuriousJoe Dec 8 '12 at 22:58
is there a way to get the right value of x, because i always mix them up and never get the right answer for x? –  CuriousJoe Dec 8 '12 at 23:04
im trying to get the smallest positive integer, but i don't always get the right answer for some odd reason –  CuriousJoe Dec 8 '12 at 23:10
ah nevermind i can do the opposite if the number is negative –  CuriousJoe Dec 8 '12 at 23:46
or i can do +m until i get a desired number –  CuriousJoe Dec 8 '12 at 23:47

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