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solve for $20x+15 \equiv 47 \pmod{4},$ if there are no solution why?

I tried to do it the following way, but i'm wondering if it is the way it should be done. Is it correct?

$$20x+15 \equiv 47 \pmod{4},$$

subtract $15$ from $47$,

$$20x \equiv 32\pmod{4},$$

divide both sides by $4$,

$$5x \equiv 8\pmod{4},$$

$$x \equiv 0.$$

311411
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hacker 21
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    $20x \equiv 32 \equiv 0 \pmod{4}$ which is trivially true no matter what $x$ is, so any $x$ is a solution. If you divide by $4$, you get $5x \equiv 8 \pmod{1}$ (don't forget to simplify the modulus) which is also true for any integer $x$ – Evariste Oct 29 '21 at 22:59
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    Your analysis is wrong because in the integers mod $4$, you’re not allowed to divide by $4 \equiv 0 \pmod 4$. – Robert Shore Oct 29 '21 at 23:04
  • Generally you can only scale (and cancel) equations )or congruences) by invertible factors if you wish to get an equivalent congruence (with same modulus) - see the linked dupes, which also explain general methods to solve linear congruences. – Bill Dubuque Oct 30 '21 at 00:37
  • Reducing mod $4,$ we get $,\color{#c00}{20}:!x\equiv_4 \color{#0a0}{32}\iff \color{#c00}0:!x\equiv_4 \color{#0a0}0$ which is true for all integers $x$. Generally you should reduce all integer arguments of sums & products (but not expts!) by the modulus since it usually simplifes modular arithmetic (this is a valid transformation due to the Congruence Sum & Product Rules). – Bill Dubuque Oct 30 '21 at 00:57

2 Answers2

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Since subtraction is a reversible step in modular arithmetic, your first and second equivalences have the same set of solutions.

Then we have $20x \equiv 32 \pmod {4} \Longleftrightarrow 4|(20x-32) \Longleftrightarrow 4|4(5x-8)$ which is trivially true for all $x$.

J. W. Tanner
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Derek Luna
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Every integer modulo $4$ is $0, 1, 2$ or $3$. Plugging $0,1,2,3$ we get $20*0+15=15\equiv 47 \mod 4$, $20*1+15=35\equiv 47 \mod 4$, $20*2+15=55\equiv 47 \mod 4$, $20*3+15=75\equiv 47 \mod 4$, so every $x$ satisfies your equation. Answer: every integer
$x$.

markvs
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