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The following is from a text about Congruences

Solve $4x\equiv 3 \pmod{19}$

The solution is $x\equiv 15\pmod{19}$

I understand till here.

Next the author writes

We can check our answer by substituting 15 into the original congruence. Is $4 ·15 \equiv 3 \pmod{19}$?

Yes, because $4 ·15 - 3 = 57 = 3 ·19$ is divisible by $19$.

I don't understand the substitution part. $x\equiv 15\pmod{19}$ and not $x = 15$. So why can $x=15$ be substituted in the original congruence to check the answer (I understand that it checks out fine).

C Squared
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user93353
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    We had the same question yesterday. Vote for closing. – Wuestenfux Aug 20 '20 at 06:48
  • @Wuestenfux - I posted a question on the same congruence yesterday. But the question itself was different. There I didn't understand how the solution was arrived at. Today, it's about how the solution is verified. – user93353 Aug 20 '20 at 07:07
  • @Wuestenfux - if you are talking about some other question & not mine from yesterday - could you please give a link? – user93353 Aug 20 '20 at 07:08
  • Here you go: https://math.stackexchange.com/questions/3796019/solving-a-congruence-cannot-understand-a-step-in-the-solution/3796022?noredirect=1#comment7820473_3796022 – Wuestenfux Aug 20 '20 at 07:11
  • @Wuestenfux - I posted that question. Both questions refer to the same congruence, but the question itself is different like I mentioned in the previous comment – user93353 Aug 20 '20 at 07:13

2 Answers2

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Suppose $\ n\ $ solves the equation $$4x\equiv 3\mod 19$$ That means $$19\mid 4n-3$$ But this implies $$19\mid 4(n+19k)-3=4n-3+4\cdot 19k$$ for every integer $\ k\ $.

Hence we can add an arbitary multiple of $\ 19\ $ to $\ n\ $ and again get a solution , hence every number congruent to $\ n\ $ modulo $\ 19\ $ is a solution.

Peter
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$15\equiv 15\mod 19$

So $x=15$ is valid in order to check.

The statement $4x\equiv 3\mod 19$ is the same as saying $4x=19q+3$ with $q\in\mathbb{Z}$. We need to find when $19q +3\equiv 0\mod 4$, because when that condition is satisfied, $x$ is an integer and $x=\frac{19q+3}{4}$.

So, when is $19q +3\equiv 0\mod 4$? Well, $3$ is always congruent to $3\mod 4$, so we need $19q$ to be congruent to $1\mod 4$. Since $19\equiv 3\mod 4$, then we need $3q+3=3(q+1)\equiv 0\mod 4$, which just means that $3(q+1)$ is divisible by $4$ with no remainder.

Testing a couple values, we find that $q=3$ works. Back-substituting, $x=\frac{19\cdot 3+3}{4}=\frac{60}{4}=15$. The next value that works is $q=7$, which yields $x=34$. Any $q=3+4k$ will work, where $k\in\mathbb{Z}$. Further substitution leads to $$4x=4(19k+15)\Longrightarrow x=19k+15\Longrightarrow x\equiv 15\mod 19$$

C Squared
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  • How does x = 15 is valid to check follow from 15≡15(mod 19)? – user93353 Aug 20 '20 at 08:40
  • any $x$ such that $x\equiv 15\mod 19$ works. He could have chosen another number, say $34$. This would have worked as well because it satisfies $34\equiv 15\mod 19$. – C Squared Aug 20 '20 at 08:44