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Here's is my approach:
$a=q_1n+1$
$b=q_2n+1$
$ab = q_1q_2n^2+q_1n+q_2n+1=(q_1q_2n+q_1+q_2)n+1$

I'm not sure if this is sufficient, and if so, whether there's a better proof.

Mel
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    This is indeed correct. I do not know a "better" proof involving less calculation. But there might be. – Cornman Oct 24 '17 at 19:28
  • Modular arithmetic is a good and powerful tool. Yes, if $a\equiv 1 \pmod n$ and $b\equiv 1 \pmod n$ then $ab\equiv 1 \pmod n$. – lulu Oct 24 '17 at 19:33

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Yes this is good. One could also use modular arithmetic to say that $a\equiv 1\pmod n$ and $b\equiv 1\pmod n$ gives $ab\equiv 1\cdot 1\equiv 1\pmod n$, however I like your approach better because it uses the actual remainder as in the division algorithm, rather than using the relationship between remainders and modular arithmetic.

Dave
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    This is a comment not answer – Leox Oct 24 '17 at 19:33
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    @Leox I provide verification of the OP's proof (i.e. I address the fact that the proof is sufficient), and I offer another proof at their request (the idea of a better proof is maybe subjective). What else should an answer to this question entail? – Dave Oct 24 '17 at 19:35
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$(ab-1)=\underbrace{(a-1)b}_\text{divisible by n}+\underbrace{(b-1)}_\text{divisible by n}$


But it is not much different from your proof :
rewrite it $(a-1)(b-1)+(a-1)+(b-1)$ and replace $(a-1)$ by $q_1n$ and same for $b$.

zwim
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