Mixing
If $n$ is an integer , find all the possible values for $(8n+6,6n+3)$
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I have got 2 questions which I could not solve:
1) if $n$ is an integer , find all the possible values for $(8n+6,6n+3)$
2)if $n$ is an integer, find all possible values of $(2n^2+3n+5,n^2+n+1)$
abstract-algebra algebra-precalculus
edited 2 days ago
greedoid
34.3k114488
asked 2 days ago
ten1o
1335
add a comment |
up vote
1
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favorite
I have got 2 questions which I could not solve:
1) if $n$ is an integer , find all the possible values for $(8n+6,6n+3)$
2)if $n$ is an integer, find all possible values of $(2n^2+3n+5,n^2+n+1)$
abstract-algebra algebra-precalculus
edited 2 days ago
greedoid
34.3k114488
asked 2 days ago
ten1o
1335
Any thoughts? An easy thing to do is to write out the values for the first few $n$, see if that generates any ideas.
– lulu
2 days ago
When I first read the question, I thought the parentheses were representing ordered pairs, but the answers are talking about gcd. Are the parentheses supposed to represent gcd? Ideal generated by these elements? I think the notation needs to be clearer.
– Acccumulation
2 days ago
@Acccumulation Haha yes I thiught the same thing "How epse can you possibly simplify this sequence in $R^2???$"
– Ovi
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have got 2 questions which I could not solve:
1) if $n$ is an integer , find all the possible values for $(8n+6,6n+3)$
2)if $n$ is an integer, find all possible values of $(2n^2+3n+5,n^2+n+1)$
abstract-algebra algebra-precalculus
edited 2 days ago
greedoid
34.3k114488
asked 2 days ago
ten1o
1335
I have got 2 questions which I could not solve:
1) if $n$ is an integer , find all the possible values for $(8n+6,6n+3)$
2)if $n$ is an integer, find all possible values of $(2n^2+3n+5,n^2+n+1)$
abstract-algebra algebra-precalculus
abstract-algebra algebra-precalculus
edited 2 days ago
greedoid
34.3k114488
asked 2 days ago
ten1o
1335
edited 2 days ago
greedoid
34.3k114488
asked 2 days ago
ten1o
1335
edited 2 days ago
greedoid
34.3k114488
edited 2 days ago
greedoid
34.3k114488
edited 2 days ago
greedoid
34.3k114488
34.3k114488
asked 2 days ago
ten1o
1335
asked 2 days ago
ten1o
1335
asked 2 days ago
ten1o
1335
1335
Any thoughts? An easy thing to do is to write out the values for the first few $n$, see if that generates any ideas.
– lulu
2 days ago
When I first read the question, I thought the parentheses were representing ordered pairs, but the answers are talking about gcd. Are the parentheses supposed to represent gcd? Ideal generated by these elements? I think the notation needs to be clearer.
– Acccumulation
2 days ago
@Acccumulation Haha yes I thiught the same thing "How epse can you possibly simplify this sequence in $R^2???$"
– Ovi
yesterday
add a comment |
Any thoughts? An easy thing to do is to write out the values for the first few $n$, see if that generates any ideas.
– lulu
2 days ago
When I first read the question, I thought the parentheses were representing ordered pairs, but the answers are talking about gcd. Are the parentheses supposed to represent gcd? Ideal generated by these elements? I think the notation needs to be clearer.
– Acccumulation
2 days ago
@Acccumulation Haha yes I thiught the same thing "How epse can you possibly simplify this sequence in $R^2???$"
– Ovi
yesterday
Any thoughts? An easy thing to do is to write out the values for the first few $n$, see if that generates any ideas.
– lulu
2 days ago
Any thoughts? An easy thing to do is to write out the values for the first few $n$, see if that generates any ideas.
– lulu
2 days ago
When I first read the question, I thought the parentheses were representing ordered pairs, but the answers are talking about gcd. Are the parentheses supposed to represent gcd? Ideal generated by these elements? I think the notation needs to be clearer.
– Acccumulation
2 days ago
When I first read the question, I thought the parentheses were representing ordered pairs, but the answers are talking about gcd. Are the parentheses supposed to represent gcd? Ideal generated by these elements? I think the notation needs to be clearer.
– Acccumulation
2 days ago
@Acccumulation Haha yes I thiught the same thing "How epse can you possibly simplify this sequence in $R^2???$"
– Ovi
yesterday
@Acccumulation Haha yes I thiught the same thing "How epse can you possibly simplify this sequence in $R^2???$"
– Ovi
yesterday
add a comment |
2 Answers
2
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oldest
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up vote
1
down vote
Let $d=gcd(8n+6,6n+3)$, then $$dmid 8n+6$$
$$dmid 6n+3$$
so $$dmid 6(8n+6)-8(6n+3)= 12$$
so $din {1,2,3,4,6,12}$ Since $6n+3$ is odd $d$ can not be $2,4,6$ or $12$ so $d=1$ or $d=3$ (which is realised at $n=3k$ for some integer $k$)
For second one:
Let $d=gcd(2n^2+3n+5,n^2 + n+1)$, then $$dmid 2n^2+3n+5$$
$$dmid n^2+n+1$$
so $$dmid 2n^2+3n+5-2(n^2 + n+1) =n+3$$
then $$dmid (n^2+n+1)-(n^2-9)-(n+3)=7$$
So $d=1$ which is ok or $d=7$ which is realised if $n=7k+4$.
answered 2 days ago
greedoid
34.3k114488
2
Should be $8n+6$ and you can multiply by $3$ and $-4$ rather than $6$ and $-8$, but good method all the same
– Mark Bennet
2 days ago
add a comment |
up vote
1
down vote
$(1)$ A euclidean sequence is $ overbrace{8n!+!6,,6n!+!3,,2n!+!3,,{-}color{#c00}6}^{Large a_{k-1} -, j a_k = a_{k+1}},$ so the gcd is
$$(2n!+!3,,color{#c00}{2cdot 3}) = (2n!+!3,color{#c00}2)(2n!+!3,color{#c00}3) = (3,2)(2n,3) = (n,3)qquadqquad $$
$(2)$ A euclidean sequence is $ 2n^2!+!3n!+!5,!!!!underbrace{n^2!+!n!+!1,, n!+!3,, color{#0a0}7}_{large f(n) equiv color{#0a0}{f(-3)},pmod{!n+3}}!!!!$ so the gcd $= (n!+!3,7)$
answered 2 days ago
Bill Dubuque
206k29189621
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
Let $d=gcd(8n+6,6n+3)$, then $$dmid 8n+6$$
$$dmid 6n+3$$
so $$dmid 6(8n+6)-8(6n+3)= 12$$
so $din {1,2,3,4,6,12}$ Since $6n+3$ is odd $d$ can not be $2,4,6$ or $12$ so $d=1$ or $d=3$ (which is realised at $n=3k$ for some integer $k$)
For second one:
Let $d=gcd(2n^2+3n+5,n^2 + n+1)$, then $$dmid 2n^2+3n+5$$
$$dmid n^2+n+1$$
so $$dmid 2n^2+3n+5-2(n^2 + n+1) =n+3$$
then $$dmid (n^2+n+1)-(n^2-9)-(n+3)=7$$
So $d=1$ which is ok or $d=7$ which is realised if $n=7k+4$.
answered 2 days ago
greedoid
34.3k114488
2
Should be $8n+6$ and you can multiply by $3$ and $-4$ rather than $6$ and $-8$, but good method all the same
– Mark Bennet
2 days ago
add a comment |
up vote
1
down vote
Let $d=gcd(8n+6,6n+3)$, then $$dmid 8n+6$$
$$dmid 6n+3$$
so $$dmid 6(8n+6)-8(6n+3)= 12$$
so $din {1,2,3,4,6,12}$ Since $6n+3$ is odd $d$ can not be $2,4,6$ or $12$ so $d=1$ or $d=3$ (which is realised at $n=3k$ for some integer $k$)
For second one:
Let $d=gcd(2n^2+3n+5,n^2 + n+1)$, then $$dmid 2n^2+3n+5$$
$$dmid n^2+n+1$$
so $$dmid 2n^2+3n+5-2(n^2 + n+1) =n+3$$
then $$dmid (n^2+n+1)-(n^2-9)-(n+3)=7$$
So $d=1$ which is ok or $d=7$ which is realised if $n=7k+4$.
answered 2 days ago
greedoid
34.3k114488
2
Should be $8n+6$ and you can multiply by $3$ and $-4$ rather than $6$ and $-8$, but good method all the same
– Mark Bennet
2 days ago
add a comment |
up vote
1
down vote
up vote
1
down vote
Let $d=gcd(8n+6,6n+3)$, then $$dmid 8n+6$$
$$dmid 6n+3$$
so $$dmid 6(8n+6)-8(6n+3)= 12$$
so $din {1,2,3,4,6,12}$ Since $6n+3$ is odd $d$ can not be $2,4,6$ or $12$ so $d=1$ or $d=3$ (which is realised at $n=3k$ for some integer $k$)
For second one:
Let $d=gcd(2n^2+3n+5,n^2 + n+1)$, then $$dmid 2n^2+3n+5$$
$$dmid n^2+n+1$$
so $$dmid 2n^2+3n+5-2(n^2 + n+1) =n+3$$
then $$dmid (n^2+n+1)-(n^2-9)-(n+3)=7$$
So $d=1$ which is ok or $d=7$ which is realised if $n=7k+4$.
answered 2 days ago
greedoid
34.3k114488
Let $d=gcd(8n+6,6n+3)$, then $$dmid 8n+6$$
$$dmid 6n+3$$
so $$dmid 6(8n+6)-8(6n+3)= 12$$
so $din {1,2,3,4,6,12}$ Since $6n+3$ is odd $d$ can not be $2,4,6$ or $12$ so $d=1$ or $d=3$ (which is realised at $n=3k$ for some integer $k$)
For second one:
Let $d=gcd(2n^2+3n+5,n^2 + n+1)$, then $$dmid 2n^2+3n+5$$
$$dmid n^2+n+1$$
so $$dmid 2n^2+3n+5-2(n^2 + n+1) =n+3$$
then $$dmid (n^2+n+1)-(n^2-9)-(n+3)=7$$
So $d=1$ which is ok or $d=7$ which is realised if $n=7k+4$.
answered 2 days ago
greedoid
34.3k114488
edited yesterday
answered 2 days ago
greedoid
34.3k114488
answered 2 days ago
greedoid
34.3k114488
answered 2 days ago
greedoid
34.3k114488
34.3k114488
2
Should be $8n+6$ and you can multiply by $3$ and $-4$ rather than $6$ and $-8$, but good method all the same
– Mark Bennet
2 days ago
add a comment |
2
Should be $8n+6$ and you can multiply by $3$ and $-4$ rather than $6$ and $-8$, but good method all the same
– Mark Bennet
2 days ago
2
2
Should be $8n+6$ and you can multiply by $3$ and $-4$ rather than $6$ and $-8$, but good method all the same
– Mark Bennet
2 days ago
Should be $8n+6$ and you can multiply by $3$ and $-4$ rather than $6$ and $-8$, but good method all the same
– Mark Bennet
2 days ago
add a comment |
up vote
1
down vote
$(1)$ A euclidean sequence is $ overbrace{8n!+!6,,6n!+!3,,2n!+!3,,{-}color{#c00}6}^{Large a_{k-1} -, j a_k = a_{k+1}},$ so the gcd is
$$(2n!+!3,,color{#c00}{2cdot 3}) = (2n!+!3,color{#c00}2)(2n!+!3,color{#c00}3) = (3,2)(2n,3) = (n,3)qquadqquad $$
$(2)$ A euclidean sequence is $ 2n^2!+!3n!+!5,!!!!underbrace{n^2!+!n!+!1,, n!+!3,, color{#0a0}7}_{large f(n) equiv color{#0a0}{f(-3)},pmod{!n+3}}!!!!$ so the gcd $= (n!+!3,7)$
answered 2 days ago
Bill Dubuque
206k29189621
add a comment |
up vote
1
down vote
$(1)$ A euclidean sequence is $ overbrace{8n!+!6,,6n!+!3,,2n!+!3,,{-}color{#c00}6}^{Large a_{k-1} -, j a_k = a_{k+1}},$ so the gcd is
$$(2n!+!3,,color{#c00}{2cdot 3}) = (2n!+!3,color{#c00}2)(2n!+!3,color{#c00}3) = (3,2)(2n,3) = (n,3)qquadqquad $$
$(2)$ A euclidean sequence is $ 2n^2!+!3n!+!5,!!!!underbrace{n^2!+!n!+!1,, n!+!3,, color{#0a0}7}_{large f(n) equiv color{#0a0}{f(-3)},pmod{!n+3}}!!!!$ so the gcd $= (n!+!3,7)$
answered 2 days ago
Bill Dubuque
206k29189621
add a comment |
up vote
1
down vote
up vote
1
down vote
$(1)$ A euclidean sequence is $ overbrace{8n!+!6,,6n!+!3,,2n!+!3,,{-}color{#c00}6}^{Large a_{k-1} -, j a_k = a_{k+1}},$ so the gcd is
$$(2n!+!3,,color{#c00}{2cdot 3}) = (2n!+!3,color{#c00}2)(2n!+!3,color{#c00}3) = (3,2)(2n,3) = (n,3)qquadqquad $$
$(2)$ A euclidean sequence is $ 2n^2!+!3n!+!5,!!!!underbrace{n^2!+!n!+!1,, n!+!3,, color{#0a0}7}_{large f(n) equiv color{#0a0}{f(-3)},pmod{!n+3}}!!!!$ so the gcd $= (n!+!3,7)$
answered 2 days ago
Bill Dubuque
206k29189621
$(1)$ A euclidean sequence is $ overbrace{8n!+!6,,6n!+!3,,2n!+!3,,{-}color{#c00}6}^{Large a_{k-1} -, j a_k = a_{k+1}},$ so the gcd is
$$(2n!+!3,,color{#c00}{2cdot 3}) = (2n!+!3,color{#c00}2)(2n!+!3,color{#c00}3) = (3,2)(2n,3) = (n,3)qquadqquad $$
$(2)$ A euclidean sequence is $ 2n^2!+!3n!+!5,!!!!underbrace{n^2!+!n!+!1,, n!+!3,, color{#0a0}7}_{large f(n) equiv color{#0a0}{f(-3)},pmod{!n+3}}!!!!$ so the gcd $= (n!+!3,7)$
answered 2 days ago
Bill Dubuque
206k29189621
edited yesterday
answered 2 days ago
Bill Dubuque
206k29189621
answered 2 days ago
Bill Dubuque
206k29189621
answered 2 days ago
Bill Dubuque
206k29189621
206k29189621
add a comment |
add a comment |
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Any thoughts? An easy thing to do is to write out the values for the first few $n$, see if that generates any ideas.
– lulu
2 days ago
When I first read the question, I thought the parentheses were representing ordered pairs, but the answers are talking about gcd. Are the parentheses supposed to represent gcd? Ideal generated by these elements? I think the notation needs to be clearer.
– Acccumulation
2 days ago
@Acccumulation Haha yes I thiught the same thing "How epse can you possibly simplify this sequence in $R^2???$"
– Ovi
yesterday