Let $a, b, c gt{1}$ be integers such that $gcd(a − 1, b − 1, c − 1) gt{1}$ Prove that $abc − 1$ is...
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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.
elementary-number-theory
asked Nov 24 at 7:45
user587054
31810
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up vote
1
down vote
favorite
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.
elementary-number-theory
asked Nov 24 at 7:45
user587054
31810
1
As $a,b,cge d+1$ so, $abc-1ge (d+1)^3-1=d^3+3d^2+3d$
– Offlaw
Nov 24 at 7:51
1
$abc - 1 = (a-1)bc + (b-1)c + (c-1)$
– achille hui
Nov 24 at 7:53
add a comment |
up vote
1
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favorite
up vote
1
down vote
favorite
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.
elementary-number-theory
asked Nov 24 at 7:45
user587054
31810
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.
elementary-number-theory
elementary-number-theory
asked Nov 24 at 7:45
user587054
31810
asked Nov 24 at 7:45
user587054
31810
asked Nov 24 at 7:45
user587054
31810
asked Nov 24 at 7:45
user587054
31810
asked Nov 24 at 7:45
user587054
31810
31810
1
As $a,b,cge d+1$ so, $abc-1ge (d+1)^3-1=d^3+3d^2+3d$
– Offlaw
Nov 24 at 7:51
1
$abc - 1 = (a-1)bc + (b-1)c + (c-1)$
– achille hui
Nov 24 at 7:53
add a comment |
1
As $a,b,cge d+1$ so, $abc-1ge (d+1)^3-1=d^3+3d^2+3d$
– Offlaw
Nov 24 at 7:51
1
$abc - 1 = (a-1)bc + (b-1)c + (c-1)$
– achille hui
Nov 24 at 7:53
1
1
As $a,b,cge d+1$ so, $abc-1ge (d+1)^3-1=d^3+3d^2+3d$
– Offlaw
Nov 24 at 7:51
As $a,b,cge d+1$ so, $abc-1ge (d+1)^3-1=d^3+3d^2+3d$
– Offlaw
Nov 24 at 7:51
1
1
$abc - 1 = (a-1)bc + (b-1)c + (c-1)$
– achille hui
Nov 24 at 7:53
$abc - 1 = (a-1)bc + (b-1)c + (c-1)$
– achille hui
Nov 24 at 7:53
add a comment |
2 Answers
2
active
oldest
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up vote
1
down vote
Note that:
$$
(md+1)(nd+1)(pd+1)-1=dtimes[m+n+p+d(mn+np+pm)+d^2mnp].
$$
Both factors are bigger than $1$.
answered Nov 24 at 7:50
yurnero
7,3701925
1
since $a,b,c,dgt1$, we know that $(m+n+p)+(mn+np+pm)d+mnpd^2ge13$
– robjohn♦
Nov 24 at 18:37
add a comment |
up vote
0
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Hint $ d,mid, abc!-!1 $ by $bmod d!: a,b,cequiv 1,Rightarrow, abcequiv 1^3equiv 1$
& $ 1 < d < abc! -! 1 $ by $ dmid a!-!1 < acolor{#c00}{bc}!-!1,$ by $,color{#c00}{bc} > 1$
answered Nov 24 at 17:43
Bill Dubuque
207k29189624
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Note that:
$$
(md+1)(nd+1)(pd+1)-1=dtimes[m+n+p+d(mn+np+pm)+d^2mnp].
$$
Both factors are bigger than $1$.
answered Nov 24 at 7:50
yurnero
7,3701925
1
since $a,b,c,dgt1$, we know that $(m+n+p)+(mn+np+pm)d+mnpd^2ge13$
– robjohn♦
Nov 24 at 18:37
add a comment |
up vote
1
down vote
Note that:
$$
(md+1)(nd+1)(pd+1)-1=dtimes[m+n+p+d(mn+np+pm)+d^2mnp].
$$
Both factors are bigger than $1$.
answered Nov 24 at 7:50
yurnero
7,3701925
1
since $a,b,c,dgt1$, we know that $(m+n+p)+(mn+np+pm)d+mnpd^2ge13$
– robjohn♦
Nov 24 at 18:37
add a comment |
up vote
1
down vote
up vote
1
down vote
Note that:
$$
(md+1)(nd+1)(pd+1)-1=dtimes[m+n+p+d(mn+np+pm)+d^2mnp].
$$
Both factors are bigger than $1$.
answered Nov 24 at 7:50
yurnero
7,3701925
Note that:
$$
(md+1)(nd+1)(pd+1)-1=dtimes[m+n+p+d(mn+np+pm)+d^2mnp].
$$
Both factors are bigger than $1$.
answered Nov 24 at 7:50
yurnero
7,3701925
answered Nov 24 at 7:50
yurnero
7,3701925
answered Nov 24 at 7:50
yurnero
7,3701925
answered Nov 24 at 7:50
yurnero
7,3701925
7,3701925
1
since $a,b,c,dgt1$, we know that $(m+n+p)+(mn+np+pm)d+mnpd^2ge13$
– robjohn♦
Nov 24 at 18:37
add a comment |
1
since $a,b,c,dgt1$, we know that $(m+n+p)+(mn+np+pm)d+mnpd^2ge13$
– robjohn♦
Nov 24 at 18:37
1
1
since $a,b,c,dgt1$, we know that $(m+n+p)+(mn+np+pm)d+mnpd^2ge13$
– robjohn♦
Nov 24 at 18:37
since $a,b,c,dgt1$, we know that $(m+n+p)+(mn+np+pm)d+mnpd^2ge13$
– robjohn♦
Nov 24 at 18:37
add a comment |
up vote
0
down vote
Hint $ d,mid, abc!-!1 $ by $bmod d!: a,b,cequiv 1,Rightarrow, abcequiv 1^3equiv 1$
& $ 1 < d < abc! -! 1 $ by $ dmid a!-!1 < acolor{#c00}{bc}!-!1,$ by $,color{#c00}{bc} > 1$
answered Nov 24 at 17:43
Bill Dubuque
207k29189624
add a comment |
up vote
0
down vote
Hint $ d,mid, abc!-!1 $ by $bmod d!: a,b,cequiv 1,Rightarrow, abcequiv 1^3equiv 1$
& $ 1 < d < abc! -! 1 $ by $ dmid a!-!1 < acolor{#c00}{bc}!-!1,$ by $,color{#c00}{bc} > 1$
answered Nov 24 at 17:43
Bill Dubuque
207k29189624
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint $ d,mid, abc!-!1 $ by $bmod d!: a,b,cequiv 1,Rightarrow, abcequiv 1^3equiv 1$
& $ 1 < d < abc! -! 1 $ by $ dmid a!-!1 < acolor{#c00}{bc}!-!1,$ by $,color{#c00}{bc} > 1$
answered Nov 24 at 17:43
Bill Dubuque
207k29189624
Hint $ d,mid, abc!-!1 $ by $bmod d!: a,b,cequiv 1,Rightarrow, abcequiv 1^3equiv 1$
& $ 1 < d < abc! -! 1 $ by $ dmid a!-!1 < acolor{#c00}{bc}!-!1,$ by $,color{#c00}{bc} > 1$
answered Nov 24 at 17:43
Bill Dubuque
207k29189624
edited Nov 24 at 19:10
answered Nov 24 at 17:43
Bill Dubuque
207k29189624
answered Nov 24 at 17:43
Bill Dubuque
207k29189624
answered Nov 24 at 17:43
Bill Dubuque
207k29189624
207k29189624
add a comment |
add a comment |
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1
As $a,b,cge d+1$ so, $abc-1ge (d+1)^3-1=d^3+3d^2+3d$
– Offlaw
Nov 24 at 7:51
1
$abc - 1 = (a-1)bc + (b-1)c + (c-1)$
– achille hui
Nov 24 at 7:53