Find all integers $a$ that satisfy $c equiv a pmod{ab+1}$












5














Let $a,b,c$ positive integers with $b|c$ I am looking for triplets that satisfy $c equiv a pmod{ab+1}$.



What I found by now is that given a solution, we can write $c=q(ab+1)+a=a(qb+1)+q$
So we get $c equiv q pmod{qb+1}$ where q is the whole part of $frac{c}{ab+1}$.



Another thing I observed is that both $a$ and $c$ are coprime to $ab+1$. This means they are both elements of the multiplicative group modulo $ab+1$.
It seems like something can be done to get more information, but I am stuck.



Thanks






number-theory modular-arithmetic







share|cite|improve this question
























  • You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
    – Shaun
    Dec 26 '18 at 16:49






  • 1




    @Shaun I have edited my question, I get what you are saying. Thanks for the advice, I hope this is ok now.
    – oren1
    Dec 26 '18 at 17:00










  • You're welcome (+1). Yeah, it's okay now.
    – Shaun
    Dec 26 '18 at 17:04










  • Help me understand, for given $b,c$ ; are you trying to find all possible values of $a$ ?
    – Haran
    Dec 26 '18 at 17:14












  • You are trying to find all solution for $a$ in the function of $a$ and $b$????
    – greedoid
    Dec 26 '18 at 17:15
















5














Let $a,b,c$ positive integers with $b|c$ I am looking for triplets that satisfy $c equiv a pmod{ab+1}$.



What I found by now is that given a solution, we can write $c=q(ab+1)+a=a(qb+1)+q$
So we get $c equiv q pmod{qb+1}$ where q is the whole part of $frac{c}{ab+1}$.



Another thing I observed is that both $a$ and $c$ are coprime to $ab+1$. This means they are both elements of the multiplicative group modulo $ab+1$.
It seems like something can be done to get more information, but I am stuck.



Thanks






number-theory modular-arithmetic







share|cite|improve this question
























  • You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
    – Shaun
    Dec 26 '18 at 16:49






  • 1




    @Shaun I have edited my question, I get what you are saying. Thanks for the advice, I hope this is ok now.
    – oren1
    Dec 26 '18 at 17:00










  • You're welcome (+1). Yeah, it's okay now.
    – Shaun
    Dec 26 '18 at 17:04










  • Help me understand, for given $b,c$ ; are you trying to find all possible values of $a$ ?
    – Haran
    Dec 26 '18 at 17:14












  • You are trying to find all solution for $a$ in the function of $a$ and $b$????
    – greedoid
    Dec 26 '18 at 17:15














5












5








5


1





Let $a,b,c$ positive integers with $b|c$ I am looking for triplets that satisfy $c equiv a pmod{ab+1}$.



What I found by now is that given a solution, we can write $c=q(ab+1)+a=a(qb+1)+q$
So we get $c equiv q pmod{qb+1}$ where q is the whole part of $frac{c}{ab+1}$.



Another thing I observed is that both $a$ and $c$ are coprime to $ab+1$. This means they are both elements of the multiplicative group modulo $ab+1$.
It seems like something can be done to get more information, but I am stuck.



Thanks






number-theory modular-arithmetic







share|cite|improve this question















Let $a,b,c$ positive integers with $b|c$ I am looking for triplets that satisfy $c equiv a pmod{ab+1}$.



What I found by now is that given a solution, we can write $c=q(ab+1)+a=a(qb+1)+q$
So we get $c equiv q pmod{qb+1}$ where q is the whole part of $frac{c}{ab+1}$.



Another thing I observed is that both $a$ and $c$ are coprime to $ab+1$. This means they are both elements of the multiplicative group modulo $ab+1$.
It seems like something can be done to get more information, but I am stuck.



Thanks






number-theory modular-arithmetic




number-theory modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 17:19

























asked Dec 26 '18 at 16:47









oren1

737




737












  • You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
    – Shaun
    Dec 26 '18 at 16:49






  • 1




    @Shaun I have edited my question, I get what you are saying. Thanks for the advice, I hope this is ok now.
    – oren1
    Dec 26 '18 at 17:00










  • You're welcome (+1). Yeah, it's okay now.
    – Shaun
    Dec 26 '18 at 17:04










  • Help me understand, for given $b,c$ ; are you trying to find all possible values of $a$ ?
    – Haran
    Dec 26 '18 at 17:14












  • You are trying to find all solution for $a$ in the function of $a$ and $b$????
    – greedoid
    Dec 26 '18 at 17:15


















  • You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
    – Shaun
    Dec 26 '18 at 16:49






  • 1




    @Shaun I have edited my question, I get what you are saying. Thanks for the advice, I hope this is ok now.
    – oren1
    Dec 26 '18 at 17:00










  • You're welcome (+1). Yeah, it's okay now.
    – Shaun
    Dec 26 '18 at 17:04










  • Help me understand, for given $b,c$ ; are you trying to find all possible values of $a$ ?
    – Haran
    Dec 26 '18 at 17:14












  • You are trying to find all solution for $a$ in the function of $a$ and $b$????
    – greedoid
    Dec 26 '18 at 17:15
















You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
– Shaun
Dec 26 '18 at 16:49




You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
– Shaun
Dec 26 '18 at 16:49




1




1




@Shaun I have edited my question, I get what you are saying. Thanks for the advice, I hope this is ok now.
– oren1
Dec 26 '18 at 17:00




@Shaun I have edited my question, I get what you are saying. Thanks for the advice, I hope this is ok now.
– oren1
Dec 26 '18 at 17:00












You're welcome (+1). Yeah, it's okay now.
– Shaun
Dec 26 '18 at 17:04




You're welcome (+1). Yeah, it's okay now.
– Shaun
Dec 26 '18 at 17:04












Help me understand, for given $b,c$ ; are you trying to find all possible values of $a$ ?
– Haran
Dec 26 '18 at 17:14






Help me understand, for given $b,c$ ; are you trying to find all possible values of $a$ ?
– Haran
Dec 26 '18 at 17:14














You are trying to find all solution for $a$ in the function of $a$ and $b$????
– greedoid
Dec 26 '18 at 17:15




You are trying to find all solution for $a$ in the function of $a$ and $b$????
– greedoid
Dec 26 '18 at 17:15










1 Answer
1






active

oldest

votes


















1














First, we have $b mid c$, which means that we can replace $c = bk$. Now, we have $a equiv bk pmod{ab+1}$. As you noted, we can see that $gcd(b,ab+1)=1$. Thus, we can instead write $k equiv frac{a}{b} pmod{ab+1}$.



We can also note that- $$ab equiv -1 pmod{ab+1} implies frac{1}{b} equiv -a pmod{ab+1} implies k equiv frac{a}{b} equiv -a^2 pmod{ab+1}$$
Thus, $k = (ab+1)q - a^2$ . Substituting this back, we get our solutions:
$${a,b,c} = {a space ,space b space, space b ((ab+1)q-a^2)}$$
where $q$ is any integer.



If you want $a$ instead in terms of $b,c$ , we can see that $c = -a^2+b^2a+bq$
$$a^2-(b^2)a+(c-bq)=0$$
This provides us a quadratic equation.
$$a = frac{b^2 pm sqrt{b^4+4bq-4c}}{2}$$
where $q$ is a suitable integer such that $b^4+4bq-4c=x^2$






share|cite|improve this answer























  • Does this answer your question? I have provided the general solution.
    – Haran
    Dec 26 '18 at 17:34










  • Yes indeed. thanks!
    – oren1
    Dec 26 '18 at 17:51











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1














First, we have $b mid c$, which means that we can replace $c = bk$. Now, we have $a equiv bk pmod{ab+1}$. As you noted, we can see that $gcd(b,ab+1)=1$. Thus, we can instead write $k equiv frac{a}{b} pmod{ab+1}$.



We can also note that- $$ab equiv -1 pmod{ab+1} implies frac{1}{b} equiv -a pmod{ab+1} implies k equiv frac{a}{b} equiv -a^2 pmod{ab+1}$$
Thus, $k = (ab+1)q - a^2$ . Substituting this back, we get our solutions:
$${a,b,c} = {a space ,space b space, space b ((ab+1)q-a^2)}$$
where $q$ is any integer.



If you want $a$ instead in terms of $b,c$ , we can see that $c = -a^2+b^2a+bq$
$$a^2-(b^2)a+(c-bq)=0$$
This provides us a quadratic equation.
$$a = frac{b^2 pm sqrt{b^4+4bq-4c}}{2}$$
where $q$ is a suitable integer such that $b^4+4bq-4c=x^2$






share|cite|improve this answer























  • Does this answer your question? I have provided the general solution.
    – Haran
    Dec 26 '18 at 17:34










  • Yes indeed. thanks!
    – oren1
    Dec 26 '18 at 17:51
















1














First, we have $b mid c$, which means that we can replace $c = bk$. Now, we have $a equiv bk pmod{ab+1}$. As you noted, we can see that $gcd(b,ab+1)=1$. Thus, we can instead write $k equiv frac{a}{b} pmod{ab+1}$.



We can also note that- $$ab equiv -1 pmod{ab+1} implies frac{1}{b} equiv -a pmod{ab+1} implies k equiv frac{a}{b} equiv -a^2 pmod{ab+1}$$
Thus, $k = (ab+1)q - a^2$ . Substituting this back, we get our solutions:
$${a,b,c} = {a space ,space b space, space b ((ab+1)q-a^2)}$$
where $q$ is any integer.



If you want $a$ instead in terms of $b,c$ , we can see that $c = -a^2+b^2a+bq$
$$a^2-(b^2)a+(c-bq)=0$$
This provides us a quadratic equation.
$$a = frac{b^2 pm sqrt{b^4+4bq-4c}}{2}$$
where $q$ is a suitable integer such that $b^4+4bq-4c=x^2$






share|cite|improve this answer























  • Does this answer your question? I have provided the general solution.
    – Haran
    Dec 26 '18 at 17:34










  • Yes indeed. thanks!
    – oren1
    Dec 26 '18 at 17:51














1












1








1






First, we have $b mid c$, which means that we can replace $c = bk$. Now, we have $a equiv bk pmod{ab+1}$. As you noted, we can see that $gcd(b,ab+1)=1$. Thus, we can instead write $k equiv frac{a}{b} pmod{ab+1}$.



We can also note that- $$ab equiv -1 pmod{ab+1} implies frac{1}{b} equiv -a pmod{ab+1} implies k equiv frac{a}{b} equiv -a^2 pmod{ab+1}$$
Thus, $k = (ab+1)q - a^2$ . Substituting this back, we get our solutions:
$${a,b,c} = {a space ,space b space, space b ((ab+1)q-a^2)}$$
where $q$ is any integer.



If you want $a$ instead in terms of $b,c$ , we can see that $c = -a^2+b^2a+bq$
$$a^2-(b^2)a+(c-bq)=0$$
This provides us a quadratic equation.
$$a = frac{b^2 pm sqrt{b^4+4bq-4c}}{2}$$
where $q$ is a suitable integer such that $b^4+4bq-4c=x^2$






share|cite|improve this answer














First, we have $b mid c$, which means that we can replace $c = bk$. Now, we have $a equiv bk pmod{ab+1}$. As you noted, we can see that $gcd(b,ab+1)=1$. Thus, we can instead write $k equiv frac{a}{b} pmod{ab+1}$.



We can also note that- $$ab equiv -1 pmod{ab+1} implies frac{1}{b} equiv -a pmod{ab+1} implies k equiv frac{a}{b} equiv -a^2 pmod{ab+1}$$
Thus, $k = (ab+1)q - a^2$ . Substituting this back, we get our solutions:
$${a,b,c} = {a space ,space b space, space b ((ab+1)q-a^2)}$$
where $q$ is any integer.



If you want $a$ instead in terms of $b,c$ , we can see that $c = -a^2+b^2a+bq$
$$a^2-(b^2)a+(c-bq)=0$$
This provides us a quadratic equation.
$$a = frac{b^2 pm sqrt{b^4+4bq-4c}}{2}$$
where $q$ is a suitable integer such that $b^4+4bq-4c=x^2$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 26 '18 at 17:48

























answered Dec 26 '18 at 17:30









Haran

788322




788322












  • Does this answer your question? I have provided the general solution.
    – Haran
    Dec 26 '18 at 17:34










  • Yes indeed. thanks!
    – oren1
    Dec 26 '18 at 17:51


















  • Does this answer your question? I have provided the general solution.
    – Haran
    Dec 26 '18 at 17:34










  • Yes indeed. thanks!
    – oren1
    Dec 26 '18 at 17:51
















Does this answer your question? I have provided the general solution.
– Haran
Dec 26 '18 at 17:34




Does this answer your question? I have provided the general solution.
– Haran
Dec 26 '18 at 17:34












Yes indeed. thanks!
– oren1
Dec 26 '18 at 17:51




Yes indeed. thanks!
– oren1
Dec 26 '18 at 17:51


















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