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I was trying to solve a problem where $\frac{n^2(n+1)^2}{4}$ has remainder $17$ when divided by $n+5$. So, I expressed it as $\frac{n^2(n+1)^2}{4} = (n+5) q + 17$. Then, multiplying four on both sides, you get $n^2(n+1)^2 = 4(n+5)q + 68$. Now, I understand that if I take this equation $\mod{n+5}$, we obtain $n^2(n+1)^2 \equiv 68 \pmod{n+5}$. However, there will be extraneous solutions since our quotient $q$ must be divisible by four. However, instead of taking our equation $\mod{n+5}$, what if we take it $\mod{4n+20}$? Then, there is no restriction on $q$, and thus (?) no extraneous solutions. Solving, we have

\begin{align*} n^2(n+1)^2 &\equiv 68 \pmod{4n + 20} \\ (4n)^2(4n+4)^2 &\equiv 68*4^2*4^2 \pmod{4n+20} \\ (-20)^2(-16)^2 &\equiv 17408 \pmod{4n+20} \\ \text{Then, dividing 4 on both }&\text{sides as well as $4n+20$ since $4 \mid 4n+20$} \\ 25600 &\equiv 4352 \pmod{n+5} \\ 21248 &\equiv 0 \pmod{n+5} \\ \Rightarrow n+5 &\mid 21248 \end{align*}

However, testing values of $n$ show that most are extraneous. What went wrong? What condition was implicitly placed that makes some of these values extraneous?

abc
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  • Same answer as in this recent exact duplicate, i.e. in your second displayed congruence, scaling by the noninvertible $4^4\pmod{4(n!+!5)},$ is not an invertible transformation $!\bmod 4(n!+!5),$ so it introduces extraneous roots. To remedy that you need to also scale the modulus too - as you did originally when scaling the original congruence by $4!:,,$ going from $!\bmod n!+!5$ to $!\bmod 4(n!+!5)$ – Bill Dubuque Nov 22 '23 at 18:46
  • @BillDubuque Ah, so what you're saying is that it doesn't matter whether there's a denominator or not, multiplying by noninvertible numbers will most likely result in extraneous solutions. Just to clarify, would solving the modular congruence when not scaling the modulus give me all the correct solutions and not leave any out (as well as of course including some extraneous solutions)? So, if I wanted to solve this congruence, I would only have to multiply 4, take $\mod{n+5}$, and test the solutions, right? – abc Nov 22 '23 at 20:00
  • Yes, $,f(r)\equiv 0,\Rightarrow, cf(r)\equiv 0,$ so every root $r$ of $f(x)$ is a root of $,cf(x)$ but the converse fails if $,c,$ is not invertible (shares factors with the modulus) $\ \ $ – Bill Dubuque Nov 22 '23 at 22:27
  • e.g. scaling $,f(x)\equiv 0\pmod{!4n},$ by $,4,$ we get $$4f(x)\equiv 0!!!\pmod{!4n}!\iff! f(x)\equiv 0!!!\pmod{!n}!\iff! f(x)\equiv \color{#0a0}0,\color{#c00}{n,2n,3n}!!!\pmod{!4n}$$ So $!\bmod 4n,$ the roots of the scaled congruence $,4f(x)\equiv 0,$ consists of the roots of the original congruence $,f(x)\equiv \color{#0a0}0,$ and also the $\rm\color{#c00}{(extraneous)}$ roots of $,f(x)\equiv a,\ ,a \in {\color{#c00}{n,2n,3n}}\ \ $ – Bill Dubuque Nov 22 '23 at 23:45

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