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Question. See the title.

Attempt. I can simplify the congruence $25 \cdot 6^{-1}\equiv 25 \cdot (-4) \; \mathrm{mod} \: 5^4$ equivalently to $6^{-1}\equiv -4 \; \mathrm{mod} \: 5^2$ which is true if $6^{-1}$ is the inverse of $6$ under $\mathrm{mod} \; 5^2$ but not true under $\mathrm{mod} \; 5^4$. However, the congruence in the question holds: $6^{-1}\equiv 521\;\mathrm{mod}\; 5^4$, and $25\cdot 521 = 13025 \equiv -100 \; \mathrm{mod} \; 5^4$. So what happened? I cannot figure it out. Thanks.

Update. I just realized that my core confusion was why $6^{-1}$ under $\mathrm{mod}\;5^4$ stays the same when we go down to $\mathrm{mod}\;5^2$, but it is actually obviously true because a multiple of $5^4$ is a multiple of $5^2$.

IncredibleSimon
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    why is it not true? $521\equiv -4 \mod 25$ – chirico Feb 11 '25 at 15:03
  • Jeez, you are right. @chirico – IncredibleSimon Feb 11 '25 at 15:05
  • Either employ the linked congruence scaling law proved in the first dupe, or use the fractional extension of the mod distributive law in the 2nd dupe to factor out $,\color{#c00}n,$ from the $!\bmod,$ (OP is $,n = 5^2,$ below) $$(6,5^2n)=1,\Rightarrow, \dfrac{\color{#c00}n}6!\bmod 5^{2}\color{#c00}n\equiv \color{#c00}n\left[\dfrac{1}6\bmod 5^2\right]\equiv n[-4],\pmod{!5^2n}\qquad$$ – Bill Dubuque Feb 11 '25 at 17:56

1 Answers1

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Division Rule

To answer your doubt, one must acknowledge the division rule for modular congruency. Given that $ac \equiv bc \pmod n$, it does not necessarily hold that $a \equiv b\pmod n$. Rather: $$ac \equiv bc \pmod n \iff a \equiv b\ \left(\text{mod }\dfrac{n}{\gcd(n, c)}\right)$$ We already see (in your example) that $521 \equiv -4\pmod{5^4}$ is wrong, but $521 \equiv -4\pmod{5^2}$ is true.

Another example is: $14 \equiv 4\pmod{10}$ does not imply $7 \equiv 2\pmod{10}$, but it correctly suggests that $7 \equiv 2\pmod 5$.

Proof

The proof of the division rule stated above is as follows. By definition of modular arithmetic, $n\ |\ c(a - b)$. WLOG, assume that $n = N\gcd(n, c)$ and $c = C\gcd(n, c)$. Thus, $N\ |\ C(a - b)$.

Since $C$ is relatively prime to $N$, we know that $N\ |\ a - b$. Therefore: $$\boxed{a \equiv b\ \left(\text{mod }\dfrac{n}{\gcd(n, c)}\right)}$$ It was also shown above that the converse is true, and its proof is literally the reverse of this process. $\blacksquare$

Joyneel Bepari
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