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Let $a, b, c \gt{1}$ be integers such that gcd$(a − 1, b − 1, c − 1) \gt{1}$ Prove that $abc − 1$ is not a prime

I have been trying to tackle this question for some time and I got stuck multiple. So far I denoted that gcd$(a − 1, b − 1, c − 1)=d$ $$\therefore a-1 \equiv 0 \pmod d, b-1 \equiv 0 \pmod d, c-1 \equiv 0 \pmod d$$ $$\therefore a \equiv 1 \pmod d, b \equiv 1 \pmod d, c \equiv 1 \pmod d$$ $$\therefore abc \equiv 1 \pmod d \iff abc-1 \equiv 0 \pmod d$$ However, I realized that $abc-1$ can still be prime if $d$ was prime and $abc-1=d$ So I attempted to assume contradiction and that gcd$(a − 1, b − 1, c − 1)=abc-1$ but failed to acheive anything. Is there a way I can carry on from here. Thank you anyways.

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    $\begingroup$ As $a,b,c\ge d+1$ so, $abc-1\ge (d+1)^3-1=d^3+3d^2+3d$ $\endgroup$ – Offlaw Nov 24 '18 at 7:51
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    $\begingroup$ $abc - 1 = (a-1)bc + (b-1)c + (c-1)$ $\endgroup$ – achille hui Nov 24 '18 at 7:53
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Note that: $$ (md+1)(nd+1)(pd+1)-1=d\times[m+n+p+d(mn+np+pm)+d^2mnp]. $$ Both factors are bigger than $1$.

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    $\begingroup$ since $a,b,c,d\gt1$, we know that $(m+n+p)+(mn+np+pm)d+mnpd^2\ge13$ $\endgroup$ – robjohn Nov 24 '18 at 18:37
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Hint $\ \ \ d\,\mid\, abc\!-\!1\ $ by $\bmod d\!:\ a,b,c\equiv 1\,\Rightarrow\, abc\equiv 1^3\equiv 1$

& $\ 1 < d < abc\! -\! 1\ $ by $\ d\mid a\!-\!1 < a\color{#c00}{bc}\!-\!1\,$ by $\,\color{#c00}{bc} > 1$

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