Find all integers $a$ and $b$, so that $ab$ is a factor in $a^5+b$
$begingroup$
Find all integers $a$ and $b$, so that $ab$ is a factor in $a^5+b$
I know that $b=-a^5$ is a solution, but are there any more solutions?
number-theory
edited Dec 19 '18 at 15:29
greedoid
43.2k1153105
asked Dec 17 '18 at 21:22
aggisanderaggisander
363
$endgroup$
add a comment |
$begingroup$
Find all integers $a$ and $b$, so that $ab$ is a factor in $a^5+b$
I know that $b=-a^5$ is a solution, but are there any more solutions?
number-theory
edited Dec 19 '18 at 15:29
greedoid
43.2k1153105
asked Dec 17 '18 at 21:22
aggisanderaggisander
363
$endgroup$
add a comment |
$begingroup$
Find all integers $a$ and $b$, so that $ab$ is a factor in $a^5+b$
I know that $b=-a^5$ is a solution, but are there any more solutions?
number-theory
edited Dec 19 '18 at 15:29
greedoid
43.2k1153105
asked Dec 17 '18 at 21:22
aggisanderaggisander
363
$endgroup$
Find all integers $a$ and $b$, so that $ab$ is a factor in $a^5+b$
I know that $b=-a^5$ is a solution, but are there any more solutions?
number-theory
number-theory
edited Dec 19 '18 at 15:29
greedoid
43.2k1153105
asked Dec 17 '18 at 21:22
aggisanderaggisander
363
edited Dec 19 '18 at 15:29
greedoid
43.2k1153105
asked Dec 17 '18 at 21:22
aggisanderaggisander
363
edited Dec 19 '18 at 15:29
greedoid
43.2k1153105
edited Dec 19 '18 at 15:29
greedoid
43.2k1153105
edited Dec 19 '18 at 15:29
greedoid
43.2k1153105
43.2k1153105
asked Dec 17 '18 at 21:22
aggisanderaggisander
363
asked Dec 17 '18 at 21:22
aggisanderaggisander
363
asked Dec 17 '18 at 21:22
aggisanderaggisander
363
363
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: Let's say we have $ab|a^n+b$ for some fixed $n$. Then $a|ab|a^n+b$, so $a|b$. Thus, let $b=ka$. Then, we need
$$ka^2big|a^n+ka$$
$$kabig|a^{n-1}+k$$
We can reduce this down to $ka|1+k$. Can you solve it from here, and use this to find all solutions to $ab|a^5+b$.
answered Dec 17 '18 at 21:32
Carl SchildkrautCarl Schildkraut
11.5k11441
$endgroup$
add a comment |
$begingroup$
So $$abmid a^5+b implies amid a^5+b implies amid b$$
so $b=ca$. Now we have $$acamid a^5+ca implies acmid a^4+c implies amid c$$
so $c=da$. Now we have $$adamid a^4+da implies admid a^3+d implies amid d$$
so $d=ea$. Now we have $$aeamid a^3+ea implies aemid a^2+e implies amid e$$
so $e=fa$. Now we have $$afamid a^2+fa implies afmid a+f implies amid f$$
so $f=ga$. Now we have $$agamid a+ga implies agmid 1+g implies gmid 1$$
so $g=1$ and $amid 2$ so $ain {1,-1,2,-2}$ and $b...$
answered Dec 19 '18 at 11:40
greedoidgreedoid
43.2k1153105
$endgroup$
add a comment |
$begingroup$
Hint $ abc = a^5!+!b !iff! b(ac!-!1) = a^5!iff! b = dfrac{a^5}{ac!-!1}.,$ It's reduced by $,gcd(a,ac!-!1)=1$ hence $,binBbb Ziff, ac!-!1 = pm1iff ldots$
answered Dec 17 '18 at 22:05
Bill DubuqueBill Dubuque
211k29193646
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Let's say we have $ab|a^n+b$ for some fixed $n$. Then $a|ab|a^n+b$, so $a|b$. Thus, let $b=ka$. Then, we need
$$ka^2big|a^n+ka$$
$$kabig|a^{n-1}+k$$
We can reduce this down to $ka|1+k$. Can you solve it from here, and use this to find all solutions to $ab|a^5+b$.
answered Dec 17 '18 at 21:32
Carl SchildkrautCarl Schildkraut
11.5k11441
$endgroup$
add a comment |
$begingroup$
Hint: Let's say we have $ab|a^n+b$ for some fixed $n$. Then $a|ab|a^n+b$, so $a|b$. Thus, let $b=ka$. Then, we need
$$ka^2big|a^n+ka$$
$$kabig|a^{n-1}+k$$
We can reduce this down to $ka|1+k$. Can you solve it from here, and use this to find all solutions to $ab|a^5+b$.
answered Dec 17 '18 at 21:32
Carl SchildkrautCarl Schildkraut
11.5k11441
$endgroup$
add a comment |
$begingroup$
Hint: Let's say we have $ab|a^n+b$ for some fixed $n$. Then $a|ab|a^n+b$, so $a|b$. Thus, let $b=ka$. Then, we need
$$ka^2big|a^n+ka$$
$$kabig|a^{n-1}+k$$
We can reduce this down to $ka|1+k$. Can you solve it from here, and use this to find all solutions to $ab|a^5+b$.
answered Dec 17 '18 at 21:32
Carl SchildkrautCarl Schildkraut
11.5k11441
$endgroup$
Hint: Let's say we have $ab|a^n+b$ for some fixed $n$. Then $a|ab|a^n+b$, so $a|b$. Thus, let $b=ka$. Then, we need
$$ka^2big|a^n+ka$$
$$kabig|a^{n-1}+k$$
We can reduce this down to $ka|1+k$. Can you solve it from here, and use this to find all solutions to $ab|a^5+b$.
answered Dec 17 '18 at 21:32
Carl SchildkrautCarl Schildkraut
11.5k11441
answered Dec 17 '18 at 21:32
Carl SchildkrautCarl Schildkraut
11.5k11441
answered Dec 17 '18 at 21:32
Carl SchildkrautCarl Schildkraut
11.5k11441
answered Dec 17 '18 at 21:32
Carl SchildkrautCarl Schildkraut
11.5k11441
11.5k11441
add a comment |
add a comment |
$begingroup$
So $$abmid a^5+b implies amid a^5+b implies amid b$$
so $b=ca$. Now we have $$acamid a^5+ca implies acmid a^4+c implies amid c$$
so $c=da$. Now we have $$adamid a^4+da implies admid a^3+d implies amid d$$
so $d=ea$. Now we have $$aeamid a^3+ea implies aemid a^2+e implies amid e$$
so $e=fa$. Now we have $$afamid a^2+fa implies afmid a+f implies amid f$$
so $f=ga$. Now we have $$agamid a+ga implies agmid 1+g implies gmid 1$$
so $g=1$ and $amid 2$ so $ain {1,-1,2,-2}$ and $b...$
answered Dec 19 '18 at 11:40
greedoidgreedoid
43.2k1153105
$endgroup$
add a comment |
$begingroup$
So $$abmid a^5+b implies amid a^5+b implies amid b$$
so $b=ca$. Now we have $$acamid a^5+ca implies acmid a^4+c implies amid c$$
so $c=da$. Now we have $$adamid a^4+da implies admid a^3+d implies amid d$$
so $d=ea$. Now we have $$aeamid a^3+ea implies aemid a^2+e implies amid e$$
so $e=fa$. Now we have $$afamid a^2+fa implies afmid a+f implies amid f$$
so $f=ga$. Now we have $$agamid a+ga implies agmid 1+g implies gmid 1$$
so $g=1$ and $amid 2$ so $ain {1,-1,2,-2}$ and $b...$
answered Dec 19 '18 at 11:40
greedoidgreedoid
43.2k1153105
$endgroup$
add a comment |
$begingroup$
So $$abmid a^5+b implies amid a^5+b implies amid b$$
so $b=ca$. Now we have $$acamid a^5+ca implies acmid a^4+c implies amid c$$
so $c=da$. Now we have $$adamid a^4+da implies admid a^3+d implies amid d$$
so $d=ea$. Now we have $$aeamid a^3+ea implies aemid a^2+e implies amid e$$
so $e=fa$. Now we have $$afamid a^2+fa implies afmid a+f implies amid f$$
so $f=ga$. Now we have $$agamid a+ga implies agmid 1+g implies gmid 1$$
so $g=1$ and $amid 2$ so $ain {1,-1,2,-2}$ and $b...$
answered Dec 19 '18 at 11:40
greedoidgreedoid
43.2k1153105
$endgroup$
So $$abmid a^5+b implies amid a^5+b implies amid b$$
so $b=ca$. Now we have $$acamid a^5+ca implies acmid a^4+c implies amid c$$
so $c=da$. Now we have $$adamid a^4+da implies admid a^3+d implies amid d$$
so $d=ea$. Now we have $$aeamid a^3+ea implies aemid a^2+e implies amid e$$
so $e=fa$. Now we have $$afamid a^2+fa implies afmid a+f implies amid f$$
so $f=ga$. Now we have $$agamid a+ga implies agmid 1+g implies gmid 1$$
so $g=1$ and $amid 2$ so $ain {1,-1,2,-2}$ and $b...$
answered Dec 19 '18 at 11:40
greedoidgreedoid
43.2k1153105
answered Dec 19 '18 at 11:40
greedoidgreedoid
43.2k1153105
answered Dec 19 '18 at 11:40
greedoidgreedoid
43.2k1153105
answered Dec 19 '18 at 11:40
greedoidgreedoid
43.2k1153105
43.2k1153105
add a comment |
add a comment |
$begingroup$
Hint $ abc = a^5!+!b !iff! b(ac!-!1) = a^5!iff! b = dfrac{a^5}{ac!-!1}.,$ It's reduced by $,gcd(a,ac!-!1)=1$ hence $,binBbb Ziff, ac!-!1 = pm1iff ldots$
answered Dec 17 '18 at 22:05
Bill DubuqueBill Dubuque
211k29193646
$endgroup$
add a comment |
$begingroup$
Hint $ abc = a^5!+!b !iff! b(ac!-!1) = a^5!iff! b = dfrac{a^5}{ac!-!1}.,$ It's reduced by $,gcd(a,ac!-!1)=1$ hence $,binBbb Ziff, ac!-!1 = pm1iff ldots$
answered Dec 17 '18 at 22:05
Bill DubuqueBill Dubuque
211k29193646
$endgroup$
add a comment |
$begingroup$
Hint $ abc = a^5!+!b !iff! b(ac!-!1) = a^5!iff! b = dfrac{a^5}{ac!-!1}.,$ It's reduced by $,gcd(a,ac!-!1)=1$ hence $,binBbb Ziff, ac!-!1 = pm1iff ldots$
answered Dec 17 '18 at 22:05
Bill DubuqueBill Dubuque
211k29193646
$endgroup$
Hint $ abc = a^5!+!b !iff! b(ac!-!1) = a^5!iff! b = dfrac{a^5}{ac!-!1}.,$ It's reduced by $,gcd(a,ac!-!1)=1$ hence $,binBbb Ziff, ac!-!1 = pm1iff ldots$
answered Dec 17 '18 at 22:05
Bill DubuqueBill Dubuque
211k29193646
edited Dec 17 '18 at 22:26
answered Dec 17 '18 at 22:05
Bill DubuqueBill Dubuque
211k29193646
answered Dec 17 '18 at 22:05
Bill DubuqueBill Dubuque
211k29193646
answered Dec 17 '18 at 22:05
Bill DubuqueBill Dubuque
211k29193646
211k29193646
add a comment |
add a comment |
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