# 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

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.

• As $a,b,c\ge d+1$ so, $abc-1\ge (d+1)^3-1=d^3+3d^2+3d$ – Offlaw Nov 24 '18 at 7:51
• $abc - 1 = (a-1)bc + (b-1)c + (c-1)$ – achille hui Nov 24 '18 at 7:53

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 $$1$$.
• since $a,b,c,d\gt1$, we know that $(m+n+p)+(mn+np+pm)d+mnpd^2\ge13$ – robjohn Nov 24 '18 at 18:37
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 $$\,\color{#c00}{bc} > 1$$