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This question is half of a larger question: Show that the equation $5x = 3$ modulo $20$ has no integer solution but $3x = 5$ does.

I can show that $3x = 5$ has an integer solution in $U(20)$ by using the fact that the multiplicative inverse of 3 is 7 in $U(20)$ since $(3 * 7)$ mod $20 = 21$ mod $20 = 1$, the identity of $U(20)$.

Multiplying, mod $20$, on the left, $$\begin{aligned}7*3x &= 7 * 5 \pmod{20}\\ x &= 15 \end{aligned}$$ Now, for the first part of the question, how do I show that there is no integer solution for $5x = 3$ mod $20$?

Does it suffice to say that there is no multiplicative inverse of 5 in $U(20)$, so there is no way to solve the equation?

Bill Dubuque
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Josh
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    For the last line, no. The equation $5x \equiv 10 \bmod 20$ has solutions even though $5$ fails to be a unit. Beware false converses. – Randall Sep 14 '23 at 15:38
  • Anyway, start playing. $20$ divides $5x-3$ and so $5x-3 = 20k$, and so.... [or just brute force it] – Randall Sep 14 '23 at 15:40
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    Every multiple of $5$ ends in $5$ or $0$ so... – John Douma Sep 14 '23 at 15:40
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    $5x\equiv 3\pmod{20}$ means $5x=20k+3$. Now, for every integer $x$, $5$ divides the LHS and the first term at the RHS, but $5\nmid 3$. – Kan't Sep 14 '23 at 15:59
  • $5$ is not in $U(20)$, because it is not coprime to $20$. So this is not an argument. The definition just says that $5x=20k+3$. This is obviously impossible, because $5$ doesn't divide $3$. – Dietrich Burde Sep 14 '23 at 16:23
  • By here in the linked dupe $,ax\equiv b \pmod{!n},$ is solvable $\iff d:=\gcd(a,n)\mid b.,$ Equivalently, it is solvable iff it is solvable $!\bmod d,\ $ [where it becomes $,0x\equiv b\pmod{!d},$]. $ $ Thus $,5x\equiv b\pmod{!5n},$ is solvable $\iff 5\mid b\ $ [with solution $,x\equiv b/5\pmod{!n},$], so it is not true that $,5,$ is not invertible $!\bmod 5n,$ implies it is unsolvable. – Bill Dubuque Sep 14 '23 at 20:17
  • But it is true that if $,\color{#0a0}{\gcd(a,b)!=! 1},$ then $,ax\equiv b \pmod{!n},$ is solvable $\iff \gcd(a,n)=1,,$ i.e. $,a,$ is invertible $!\bmod n,,$ since by above it is solvable $\iff \gcd(a,n)\mid\color{#c00} b\iff \gcd(a,n)=1,,$ since then $,d=\gcd(a,n),$ is a common divisor of $\rm\color{#0a0}{coprimes}$ $,a,$ and $,\color{#c00}b,,$ so $,d=1.,$ In modular fraction language: $!\bmod n!:, $ $\rm\color{#0a0}{reduced}$ fraction $,x\equiv b/a,$ exists $\iff 1/a,$ exists. See here for elaboration. – Bill Dubuque Sep 14 '23 at 20:34
  • In your example $,\color{#0a0}{a,b} = 5,3,$ are coprime, so $,\gcd(a,n)=\gcd(5,20)>1\Rightarrow$ unsolvable, i.e. your argument does suffice, but you'd need to justify this claim (e.g. as in my prior comment). – Bill Dubuque Sep 14 '23 at 20:45
  • @Dietrich Actually it is a correct inference in this case of a $\rm\color{#0a0}{reduced}$ modular fraction - see my prior $2$ comments. – Bill Dubuque Sep 14 '23 at 20:48

1 Answers1

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If you multiply both sides of the equation by 4, you get a contradiction.

Tony Dean
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