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How would one solve for $n$ in a modular system of equation(s) such as this one?

$42n+7$ $=$ $1$ $\pmod {7n+1}$

$12n+3$ $=$ $1$ $\pmod {6n+1}$

Please show work (just as in algebra) if possible. Thanks in advance.

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    The modular-forms tag involves a different class of mathematics than your question... – 2017-02-04

3 Answers 3

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Both of these are trivially true, giving no information on $n$.

$\begin{align} 42n+7 &\equiv 1\bmod {7n+1}\\ (42n+7)-6(7n+1) &\equiv 1\bmod {7n+1}\\ (42n+7)-(42n+6) &\equiv 1\bmod {7n+1}\\ 1 &\equiv 1\bmod {7n+1}\\ \end{align}$

$\begin{align} 12n+3 &\equiv 1\bmod {6n+1}\\ (12n+3)-2(6n+1) &\equiv 1\bmod {6n+1}\\ (12n+3)-(12n+2) &\equiv 1\bmod {6n+1}\\ 1 &\equiv 1\bmod {6n+1}\\ \end{align}$

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Consider \begin{align*} 42n+1 & \equiv 1 \pmod{7n+1}\\ 6(7n+1)-5 & \equiv 1 \pmod{7n+1}\\ -5 & \equiv 1 \pmod{7n+1}\\ 6 & \equiv 0 \pmod{7n+1} \end{align*} Thus $7n+1$ should divide $6$. This is possible for $n=-1,0$.

Similarly try the second congruence as well.

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    however first eqn is $42n+7\equiv 1$ – 2017-02-04
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    @Joffan OP changed the question many times. – 2017-02-04
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    I checked edit history before commenting, but could've been in grace period I guess. – 2017-02-04
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    $n$ $=$ $4$ seems to work. – 2017-02-04
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${\rm mod}\ 7n\!+\!1\!:\,\ 7n\equiv -1\,\overset{\large\times 6}\Rightarrow\, 42n\equiv-6\,\overset{\large + 6}\Rightarrow\,42n+7\equiv 1$

${\rm mod}\ 6n\!+\!1\!:\,\ 6n\equiv -1\,\overset{\large\times 2}\Rightarrow\, 12n\equiv-2\,\overset{\large + 3}\Rightarrow\,12n+3\equiv 1$