Mixing
Find conditions on $a, b, c$, and $d$ with $ane -1, 0, 1$ such that $dmid(a^n+bn+c)$ for $n ge 1$.
This is a generalization of
Using induction, show that $4^n +15n - 1$ is divisible by $9$ for all $n geq 1$
I want to find conditions on
$a, b, c$, and $d$
with
$ane -1, 0, 1$
such that
$dmid(a^n+bn+c)$
for
$n ge 1$.
Here is my result:
A sufficient condition
is that
$a+b+c ne 0$
and
all of
$a+b+c,
b(a-1)$,
and
$c(a-1)-b$
are divisible by $d$.
For the problem
that prompted this,
with
$a=4, b=15, c=-1$,
these are
$18, 45,$
and
$-18$.
sequences-and-series elementary-number-theory divisibility
edited Nov 22 '18 at 10:47
Saad
19.7k92352
asked Nov 16 '18 at 3:56
marty cohenmarty cohen
72.8k549128
add a comment |
This is a generalization of
Using induction, show that $4^n +15n - 1$ is divisible by $9$ for all $n geq 1$
I want to find conditions on
$a, b, c$, and $d$
with
$ane -1, 0, 1$
such that
$dmid(a^n+bn+c)$
for
$n ge 1$.
Here is my result:
A sufficient condition
is that
$a+b+c ne 0$
and
all of
$a+b+c,
b(a-1)$,
and
$c(a-1)-b$
are divisible by $d$.
For the problem
that prompted this,
with
$a=4, b=15, c=-1$,
these are
$18, 45,$
and
$-18$.
sequences-and-series elementary-number-theory divisibility
edited Nov 22 '18 at 10:47
Saad
19.7k92352
asked Nov 16 '18 at 3:56
marty cohenmarty cohen
72.8k549128
1
This is a dupe (of at least a couple threads)
– Bill Dubuque
Nov 16 '18 at 4:29
Wouldn't be surprised. Might even be a dupe of myself, the way my memory works. Anyway, I worked this out just today completely independently. If you find the dupe, I'll upvote you. What the heck, I'll upvote you anyway.
– marty cohen
Nov 16 '18 at 5:11
2
I found a couple, e.g. here and here. There are likely more.
– Bill Dubuque
Nov 16 '18 at 15:35
I have done this as an excercise of induction. I think it will be hard to find the conditions.
– OppoInfinity
Nov 19 '18 at 4:33
add a comment |
This is a generalization of
Using induction, show that $4^n +15n - 1$ is divisible by $9$ for all $n geq 1$
I want to find conditions on
$a, b, c$, and $d$
with
$ane -1, 0, 1$
such that
$dmid(a^n+bn+c)$
for
$n ge 1$.
Here is my result:
A sufficient condition
is that
$a+b+c ne 0$
and
all of
$a+b+c,
b(a-1)$,
and
$c(a-1)-b$
are divisible by $d$.
For the problem
that prompted this,
with
$a=4, b=15, c=-1$,
these are
$18, 45,$
and
$-18$.
sequences-and-series elementary-number-theory divisibility
edited Nov 22 '18 at 10:47
Saad
19.7k92352
asked Nov 16 '18 at 3:56
marty cohenmarty cohen
72.8k549128
This is a generalization of
Using induction, show that $4^n +15n - 1$ is divisible by $9$ for all $n geq 1$
I want to find conditions on
$a, b, c$, and $d$
with
$ane -1, 0, 1$
such that
$dmid(a^n+bn+c)$
for
$n ge 1$.
Here is my result:
A sufficient condition
is that
$a+b+c ne 0$
and
all of
$a+b+c,
b(a-1)$,
and
$c(a-1)-b$
are divisible by $d$.
For the problem
that prompted this,
with
$a=4, b=15, c=-1$,
these are
$18, 45,$
and
$-18$.
sequences-and-series elementary-number-theory divisibility
sequences-and-series elementary-number-theory divisibility
edited Nov 22 '18 at 10:47
Saad
19.7k92352
asked Nov 16 '18 at 3:56
marty cohenmarty cohen
72.8k549128
edited Nov 22 '18 at 10:47
Saad
19.7k92352
asked Nov 16 '18 at 3:56
marty cohenmarty cohen
72.8k549128
edited Nov 22 '18 at 10:47
Saad
19.7k92352
edited Nov 22 '18 at 10:47
Saad
19.7k92352
edited Nov 22 '18 at 10:47
Saad
19.7k92352
19.7k92352
asked Nov 16 '18 at 3:56
marty cohenmarty cohen
72.8k549128
asked Nov 16 '18 at 3:56
marty cohenmarty cohen
72.8k549128
asked Nov 16 '18 at 3:56
marty cohenmarty cohen
72.8k549128
72.8k549128
1
This is a dupe (of at least a couple threads)
– Bill Dubuque
Nov 16 '18 at 4:29
Wouldn't be surprised. Might even be a dupe of myself, the way my memory works. Anyway, I worked this out just today completely independently. If you find the dupe, I'll upvote you. What the heck, I'll upvote you anyway.
– marty cohen
Nov 16 '18 at 5:11
2
I found a couple, e.g. here and here. There are likely more.
– Bill Dubuque
Nov 16 '18 at 15:35
I have done this as an excercise of induction. I think it will be hard to find the conditions.
– OppoInfinity
Nov 19 '18 at 4:33
add a comment |
1
This is a dupe (of at least a couple threads)
– Bill Dubuque
Nov 16 '18 at 4:29
Wouldn't be surprised. Might even be a dupe of myself, the way my memory works. Anyway, I worked this out just today completely independently. If you find the dupe, I'll upvote you. What the heck, I'll upvote you anyway.
– marty cohen
Nov 16 '18 at 5:11
2
I found a couple, e.g. here and here. There are likely more.
– Bill Dubuque
Nov 16 '18 at 15:35
I have done this as an excercise of induction. I think it will be hard to find the conditions.
– OppoInfinity
Nov 19 '18 at 4:33
1
1
This is a dupe (of at least a couple threads)
– Bill Dubuque
Nov 16 '18 at 4:29
This is a dupe (of at least a couple threads)
– Bill Dubuque
Nov 16 '18 at 4:29
Wouldn't be surprised. Might even be a dupe of myself, the way my memory works. Anyway, I worked this out just today completely independently. If you find the dupe, I'll upvote you. What the heck, I'll upvote you anyway.
– marty cohen
Nov 16 '18 at 5:11
Wouldn't be surprised. Might even be a dupe of myself, the way my memory works. Anyway, I worked this out just today completely independently. If you find the dupe, I'll upvote you. What the heck, I'll upvote you anyway.
– marty cohen
Nov 16 '18 at 5:11
2
2
I found a couple, e.g. here and here. There are likely more.
– Bill Dubuque
Nov 16 '18 at 15:35
I found a couple, e.g. here and here. There are likely more.
– Bill Dubuque
Nov 16 '18 at 15:35
I have done this as an excercise of induction. I think it will be hard to find the conditions.
– OppoInfinity
Nov 19 '18 at 4:33
I have done this as an excercise of induction. I think it will be hard to find the conditions.
– OppoInfinity
Nov 19 '18 at 4:33
add a comment |
1 Answer
1
active
oldest
votes
First, let $$gcd(a,d)=k$$therefore $$k|d|a^n+bn+c$$since $k|a$ we obtain$$k|bn+c$$ for any $nge 1$. This means that $$k|b+c\k|2b+c$$so $$k|b\k|c$$from which we can conclude that if $(a,b,c,d)$ is an answer, then so is $$left({a',b',c',d'}right)$$where$$a'={aover gcd(a,d)}\b'={bover gcd(a,d)}\c'={cover gcd(a,d)}\d'={dover gcd(a,d)}$$which means that $gcd(a,d)$. Therefore without loss of generality, we assume $gcd(a,d)=1$ furthermore $dinBbb N$ and solve the problem, i.e. find the 4-tuple $(a,b,c,d)$. All other answers can be found as $(ka,kb,kc,kd)$ for $kin Bbb Z$.
The case where $gcd(a,d)=1$ and $dinBbb N$
We show that there exists $p$ such that $1le ple d$ such that $$a^pequiv 1mod {d}$$
proof: first off, note that $$Big{a^kmod d Big| 1le kle dBig}subseteqBig{1,2,cdots ,d Big}$$ if such a $p$ doesn't exist, then we must have $i,j>i$ for which $$a^iequiv a^jmod d$$since $gcd(a,d)=1$ we obtain $$a^{j-i}equiv 1mod d$$since $1le j-ile d$, this contradicts with our first assumption on non-existence of $p$ since we can choose $p=j-i$. This completes our proof.
Since such a $p$ exists, by substituting $nto np$ in $d|a^n+bn+c$ we conclude that$$d|a^{np}+bnp+c$$therefore $$d|bnp+c+1$$for any $ninBbb Z$. Also $$d|2bnp+c+1$$ by a simple argument we obtain $$d|c+1$$ which yields to $$d|a^n+bn-1$$for $nin Bbb Z$ and $gcd(a,d)=1$.
Minor Conclusion
We have proved so far that
(1) if $gcd(a,d)=1$ then $$d|a^n+bn+ciff d|a^n+bn-1$$
(2) the case where $gcd(a,d)ne 1$ can be reduced to the case with $gcd(a,d)=1$.
If $u$ is the smallest such that $$d|a^u-1$$ we easily conclude that $$d|bu$$ then an algorithm goes like this:
Choose $a,d$ such that $gcd(a,d)=1$ and any $c$ with $d|c+1$. Then find $1le ule d$ such that $d|a^u-1$ (such a $u$ exists and is unique as proved before). Therefore for any $b$ with $d|bu$ and every $nin Bbb Z$ we have $$d|a^n+bn+c$$
We illuminate the result with some examples:
for $a=4,b=15,c=-1$, the condition $d|c+1$ is redundant since $c+1=0$. Also $gcd(a,d)=1$ means all the odd values of $d$. Let $d=9$ then $$9|4^3-1$$ therefore $u=3$ and $$9|3b$$ which means that $b=3k$ for any $kinBbb Z$. Now let $d=45$ therefore $$45|4^6-1$$and we have $$45|6bto 15|b$$. Choosing $b=15$ satisfies both the examples.
answered Jan 5 at 9:27
Mostafa AyazMostafa Ayaz
14.5k3937
add a comment |
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First, let $$gcd(a,d)=k$$therefore $$k|d|a^n+bn+c$$since $k|a$ we obtain$$k|bn+c$$ for any $nge 1$. This means that $$k|b+c\k|2b+c$$so $$k|b\k|c$$from which we can conclude that if $(a,b,c,d)$ is an answer, then so is $$left({a',b',c',d'}right)$$where$$a'={aover gcd(a,d)}\b'={bover gcd(a,d)}\c'={cover gcd(a,d)}\d'={dover gcd(a,d)}$$which means that $gcd(a,d)$. Therefore without loss of generality, we assume $gcd(a,d)=1$ furthermore $dinBbb N$ and solve the problem, i.e. find the 4-tuple $(a,b,c,d)$. All other answers can be found as $(ka,kb,kc,kd)$ for $kin Bbb Z$.
The case where $gcd(a,d)=1$ and $dinBbb N$
We show that there exists $p$ such that $1le ple d$ such that $$a^pequiv 1mod {d}$$
proof: first off, note that $$Big{a^kmod d Big| 1le kle dBig}subseteqBig{1,2,cdots ,d Big}$$ if such a $p$ doesn't exist, then we must have $i,j>i$ for which $$a^iequiv a^jmod d$$since $gcd(a,d)=1$ we obtain $$a^{j-i}equiv 1mod d$$since $1le j-ile d$, this contradicts with our first assumption on non-existence of $p$ since we can choose $p=j-i$. This completes our proof.
Since such a $p$ exists, by substituting $nto np$ in $d|a^n+bn+c$ we conclude that$$d|a^{np}+bnp+c$$therefore $$d|bnp+c+1$$for any $ninBbb Z$. Also $$d|2bnp+c+1$$ by a simple argument we obtain $$d|c+1$$ which yields to $$d|a^n+bn-1$$for $nin Bbb Z$ and $gcd(a,d)=1$.
Minor Conclusion
We have proved so far that
(1) if $gcd(a,d)=1$ then $$d|a^n+bn+ciff d|a^n+bn-1$$
(2) the case where $gcd(a,d)ne 1$ can be reduced to the case with $gcd(a,d)=1$.
If $u$ is the smallest such that $$d|a^u-1$$ we easily conclude that $$d|bu$$ then an algorithm goes like this:
Choose $a,d$ such that $gcd(a,d)=1$ and any $c$ with $d|c+1$. Then find $1le ule d$ such that $d|a^u-1$ (such a $u$ exists and is unique as proved before). Therefore for any $b$ with $d|bu$ and every $nin Bbb Z$ we have $$d|a^n+bn+c$$
We illuminate the result with some examples:
for $a=4,b=15,c=-1$, the condition $d|c+1$ is redundant since $c+1=0$. Also $gcd(a,d)=1$ means all the odd values of $d$. Let $d=9$ then $$9|4^3-1$$ therefore $u=3$ and $$9|3b$$ which means that $b=3k$ for any $kinBbb Z$. Now let $d=45$ therefore $$45|4^6-1$$and we have $$45|6bto 15|b$$. Choosing $b=15$ satisfies both the examples.
answered Jan 5 at 9:27
Mostafa AyazMostafa Ayaz
14.5k3937
add a comment |
First, let $$gcd(a,d)=k$$therefore $$k|d|a^n+bn+c$$since $k|a$ we obtain$$k|bn+c$$ for any $nge 1$. This means that $$k|b+c\k|2b+c$$so $$k|b\k|c$$from which we can conclude that if $(a,b,c,d)$ is an answer, then so is $$left({a',b',c',d'}right)$$where$$a'={aover gcd(a,d)}\b'={bover gcd(a,d)}\c'={cover gcd(a,d)}\d'={dover gcd(a,d)}$$which means that $gcd(a,d)$. Therefore without loss of generality, we assume $gcd(a,d)=1$ furthermore $dinBbb N$ and solve the problem, i.e. find the 4-tuple $(a,b,c,d)$. All other answers can be found as $(ka,kb,kc,kd)$ for $kin Bbb Z$.
The case where $gcd(a,d)=1$ and $dinBbb N$
We show that there exists $p$ such that $1le ple d$ such that $$a^pequiv 1mod {d}$$
proof: first off, note that $$Big{a^kmod d Big| 1le kle dBig}subseteqBig{1,2,cdots ,d Big}$$ if such a $p$ doesn't exist, then we must have $i,j>i$ for which $$a^iequiv a^jmod d$$since $gcd(a,d)=1$ we obtain $$a^{j-i}equiv 1mod d$$since $1le j-ile d$, this contradicts with our first assumption on non-existence of $p$ since we can choose $p=j-i$. This completes our proof.
Since such a $p$ exists, by substituting $nto np$ in $d|a^n+bn+c$ we conclude that$$d|a^{np}+bnp+c$$therefore $$d|bnp+c+1$$for any $ninBbb Z$. Also $$d|2bnp+c+1$$ by a simple argument we obtain $$d|c+1$$ which yields to $$d|a^n+bn-1$$for $nin Bbb Z$ and $gcd(a,d)=1$.
Minor Conclusion
We have proved so far that
(1) if $gcd(a,d)=1$ then $$d|a^n+bn+ciff d|a^n+bn-1$$
(2) the case where $gcd(a,d)ne 1$ can be reduced to the case with $gcd(a,d)=1$.
If $u$ is the smallest such that $$d|a^u-1$$ we easily conclude that $$d|bu$$ then an algorithm goes like this:
Choose $a,d$ such that $gcd(a,d)=1$ and any $c$ with $d|c+1$. Then find $1le ule d$ such that $d|a^u-1$ (such a $u$ exists and is unique as proved before). Therefore for any $b$ with $d|bu$ and every $nin Bbb Z$ we have $$d|a^n+bn+c$$
We illuminate the result with some examples:
for $a=4,b=15,c=-1$, the condition $d|c+1$ is redundant since $c+1=0$. Also $gcd(a,d)=1$ means all the odd values of $d$. Let $d=9$ then $$9|4^3-1$$ therefore $u=3$ and $$9|3b$$ which means that $b=3k$ for any $kinBbb Z$. Now let $d=45$ therefore $$45|4^6-1$$and we have $$45|6bto 15|b$$. Choosing $b=15$ satisfies both the examples.
answered Jan 5 at 9:27
Mostafa AyazMostafa Ayaz
14.5k3937
add a comment |
First, let $$gcd(a,d)=k$$therefore $$k|d|a^n+bn+c$$since $k|a$ we obtain$$k|bn+c$$ for any $nge 1$. This means that $$k|b+c\k|2b+c$$so $$k|b\k|c$$from which we can conclude that if $(a,b,c,d)$ is an answer, then so is $$left({a',b',c',d'}right)$$where$$a'={aover gcd(a,d)}\b'={bover gcd(a,d)}\c'={cover gcd(a,d)}\d'={dover gcd(a,d)}$$which means that $gcd(a,d)$. Therefore without loss of generality, we assume $gcd(a,d)=1$ furthermore $dinBbb N$ and solve the problem, i.e. find the 4-tuple $(a,b,c,d)$. All other answers can be found as $(ka,kb,kc,kd)$ for $kin Bbb Z$.
The case where $gcd(a,d)=1$ and $dinBbb N$
We show that there exists $p$ such that $1le ple d$ such that $$a^pequiv 1mod {d}$$
proof: first off, note that $$Big{a^kmod d Big| 1le kle dBig}subseteqBig{1,2,cdots ,d Big}$$ if such a $p$ doesn't exist, then we must have $i,j>i$ for which $$a^iequiv a^jmod d$$since $gcd(a,d)=1$ we obtain $$a^{j-i}equiv 1mod d$$since $1le j-ile d$, this contradicts with our first assumption on non-existence of $p$ since we can choose $p=j-i$. This completes our proof.
Since such a $p$ exists, by substituting $nto np$ in $d|a^n+bn+c$ we conclude that$$d|a^{np}+bnp+c$$therefore $$d|bnp+c+1$$for any $ninBbb Z$. Also $$d|2bnp+c+1$$ by a simple argument we obtain $$d|c+1$$ which yields to $$d|a^n+bn-1$$for $nin Bbb Z$ and $gcd(a,d)=1$.
Minor Conclusion
We have proved so far that
(1) if $gcd(a,d)=1$ then $$d|a^n+bn+ciff d|a^n+bn-1$$
(2) the case where $gcd(a,d)ne 1$ can be reduced to the case with $gcd(a,d)=1$.
If $u$ is the smallest such that $$d|a^u-1$$ we easily conclude that $$d|bu$$ then an algorithm goes like this:
Choose $a,d$ such that $gcd(a,d)=1$ and any $c$ with $d|c+1$. Then find $1le ule d$ such that $d|a^u-1$ (such a $u$ exists and is unique as proved before). Therefore for any $b$ with $d|bu$ and every $nin Bbb Z$ we have $$d|a^n+bn+c$$
We illuminate the result with some examples:
for $a=4,b=15,c=-1$, the condition $d|c+1$ is redundant since $c+1=0$. Also $gcd(a,d)=1$ means all the odd values of $d$. Let $d=9$ then $$9|4^3-1$$ therefore $u=3$ and $$9|3b$$ which means that $b=3k$ for any $kinBbb Z$. Now let $d=45$ therefore $$45|4^6-1$$and we have $$45|6bto 15|b$$. Choosing $b=15$ satisfies both the examples.
answered Jan 5 at 9:27
Mostafa AyazMostafa Ayaz
14.5k3937
First, let $$gcd(a,d)=k$$therefore $$k|d|a^n+bn+c$$since $k|a$ we obtain$$k|bn+c$$ for any $nge 1$. This means that $$k|b+c\k|2b+c$$so $$k|b\k|c$$from which we can conclude that if $(a,b,c,d)$ is an answer, then so is $$left({a',b',c',d'}right)$$where$$a'={aover gcd(a,d)}\b'={bover gcd(a,d)}\c'={cover gcd(a,d)}\d'={dover gcd(a,d)}$$which means that $gcd(a,d)$. Therefore without loss of generality, we assume $gcd(a,d)=1$ furthermore $dinBbb N$ and solve the problem, i.e. find the 4-tuple $(a,b,c,d)$. All other answers can be found as $(ka,kb,kc,kd)$ for $kin Bbb Z$.
The case where $gcd(a,d)=1$ and $dinBbb N$
We show that there exists $p$ such that $1le ple d$ such that $$a^pequiv 1mod {d}$$
proof: first off, note that $$Big{a^kmod d Big| 1le kle dBig}subseteqBig{1,2,cdots ,d Big}$$ if such a $p$ doesn't exist, then we must have $i,j>i$ for which $$a^iequiv a^jmod d$$since $gcd(a,d)=1$ we obtain $$a^{j-i}equiv 1mod d$$since $1le j-ile d$, this contradicts with our first assumption on non-existence of $p$ since we can choose $p=j-i$. This completes our proof.
Since such a $p$ exists, by substituting $nto np$ in $d|a^n+bn+c$ we conclude that$$d|a^{np}+bnp+c$$therefore $$d|bnp+c+1$$for any $ninBbb Z$. Also $$d|2bnp+c+1$$ by a simple argument we obtain $$d|c+1$$ which yields to $$d|a^n+bn-1$$for $nin Bbb Z$ and $gcd(a,d)=1$.
Minor Conclusion
We have proved so far that
(1) if $gcd(a,d)=1$ then $$d|a^n+bn+ciff d|a^n+bn-1$$
(2) the case where $gcd(a,d)ne 1$ can be reduced to the case with $gcd(a,d)=1$.
If $u$ is the smallest such that $$d|a^u-1$$ we easily conclude that $$d|bu$$ then an algorithm goes like this:
Choose $a,d$ such that $gcd(a,d)=1$ and any $c$ with $d|c+1$. Then find $1le ule d$ such that $d|a^u-1$ (such a $u$ exists and is unique as proved before). Therefore for any $b$ with $d|bu$ and every $nin Bbb Z$ we have $$d|a^n+bn+c$$
We illuminate the result with some examples:
for $a=4,b=15,c=-1$, the condition $d|c+1$ is redundant since $c+1=0$. Also $gcd(a,d)=1$ means all the odd values of $d$. Let $d=9$ then $$9|4^3-1$$ therefore $u=3$ and $$9|3b$$ which means that $b=3k$ for any $kinBbb Z$. Now let $d=45$ therefore $$45|4^6-1$$and we have $$45|6bto 15|b$$. Choosing $b=15$ satisfies both the examples.
answered Jan 5 at 9:27
Mostafa AyazMostafa Ayaz
14.5k3937
answered Jan 5 at 9:27
Mostafa AyazMostafa Ayaz
14.5k3937
answered Jan 5 at 9:27
Mostafa AyazMostafa Ayaz
14.5k3937
answered Jan 5 at 9:27
Mostafa AyazMostafa Ayaz
14.5k3937
14.5k3937
add a comment |
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This is a dupe (of at least a couple threads)
– Bill Dubuque
Nov 16 '18 at 4:29
Wouldn't be surprised. Might even be a dupe of myself, the way my memory works. Anyway, I worked this out just today completely independently. If you find the dupe, I'll upvote you. What the heck, I'll upvote you anyway.
– marty cohen
Nov 16 '18 at 5:11
2
I found a couple, e.g. here and here. There are likely more.
– Bill Dubuque
Nov 16 '18 at 15:35
I have done this as an excercise of induction. I think it will be hard to find the conditions.
– OppoInfinity
Nov 19 '18 at 4:33