Proving That Consecutive Fibonacci Numbers are Relatively Prime












2












$begingroup$


The Problem:




Prove that consecutive Fibonacci numbers are relatively prime.




I have seen proofs where people use induction and show that if $gcd(F_n, F_{n+1})=1$, then $gcd(F_{n+1}, F_{n+2})=1$ through the Fibonacci property.



My proof uses induction as well, but in a different way.




We will use induction to show that if such a base case exists that $F_{n}$ and $F_{n+1}$ are both divisible by some integral $k ge 2$, then all $F_{i}$ are divisible by $k$.



$$F_{n} equiv F_{n+1} equiv 0 pmod{k}$$
$$F_{n} equiv F_{n}+F_{n-1} equiv 0 pmod{k}$$
$$F_{n} equiv F_{n-1} equiv 0 pmod{k}$$



Since $F_{3}=2$ and $F_4=3$ are not divisible be any common integer $k ge 2$, our assumption is wrong. Hence, $F_{n}$ and $F_{n+1}$ are relatively prime.







proof-verification induction fibonacci-numbers coprime







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    The Problem:




    Prove that consecutive Fibonacci numbers are relatively prime.




    I have seen proofs where people use induction and show that if $gcd(F_n, F_{n+1})=1$, then $gcd(F_{n+1}, F_{n+2})=1$ through the Fibonacci property.



    My proof uses induction as well, but in a different way.




    We will use induction to show that if such a base case exists that $F_{n}$ and $F_{n+1}$ are both divisible by some integral $k ge 2$, then all $F_{i}$ are divisible by $k$.



    $$F_{n} equiv F_{n+1} equiv 0 pmod{k}$$
    $$F_{n} equiv F_{n}+F_{n-1} equiv 0 pmod{k}$$
    $$F_{n} equiv F_{n-1} equiv 0 pmod{k}$$



    Since $F_{3}=2$ and $F_4=3$ are not divisible be any common integer $k ge 2$, our assumption is wrong. Hence, $F_{n}$ and $F_{n+1}$ are relatively prime.







    proof-verification induction fibonacci-numbers coprime







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      The Problem:




      Prove that consecutive Fibonacci numbers are relatively prime.




      I have seen proofs where people use induction and show that if $gcd(F_n, F_{n+1})=1$, then $gcd(F_{n+1}, F_{n+2})=1$ through the Fibonacci property.



      My proof uses induction as well, but in a different way.




      We will use induction to show that if such a base case exists that $F_{n}$ and $F_{n+1}$ are both divisible by some integral $k ge 2$, then all $F_{i}$ are divisible by $k$.



      $$F_{n} equiv F_{n+1} equiv 0 pmod{k}$$
      $$F_{n} equiv F_{n}+F_{n-1} equiv 0 pmod{k}$$
      $$F_{n} equiv F_{n-1} equiv 0 pmod{k}$$



      Since $F_{3}=2$ and $F_4=3$ are not divisible be any common integer $k ge 2$, our assumption is wrong. Hence, $F_{n}$ and $F_{n+1}$ are relatively prime.







      proof-verification induction fibonacci-numbers coprime







      share|cite|improve this question











      $endgroup$




      The Problem:




      Prove that consecutive Fibonacci numbers are relatively prime.




      I have seen proofs where people use induction and show that if $gcd(F_n, F_{n+1})=1$, then $gcd(F_{n+1}, F_{n+2})=1$ through the Fibonacci property.



      My proof uses induction as well, but in a different way.




      We will use induction to show that if such a base case exists that $F_{n}$ and $F_{n+1}$ are both divisible by some integral $k ge 2$, then all $F_{i}$ are divisible by $k$.



      $$F_{n} equiv F_{n+1} equiv 0 pmod{k}$$
      $$F_{n} equiv F_{n}+F_{n-1} equiv 0 pmod{k}$$
      $$F_{n} equiv F_{n-1} equiv 0 pmod{k}$$



      Since $F_{3}=2$ and $F_4=3$ are not divisible be any common integer $k ge 2$, our assumption is wrong. Hence, $F_{n}$ and $F_{n+1}$ are relatively prime.







      proof-verification induction fibonacci-numbers coprime




      proof-verification induction fibonacci-numbers coprime






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 3 at 23:11









      José Carlos Santos

      155k22124227




      155k22124227










      asked Jan 3 at 22:41









      Shrey JoshiShrey Joshi

      30713




      30713






















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          It is a waste of time to use modular arithmetic for this. What you proved was that$$kmid F_nwedge kmid F_{n+1}implies kmid F_{n-1},$$which is indeed correct. But, after having done this, I think that your best option is to say that you are using the well-ordering principle: if there was some $ninmathbb N$ such that $F_n$ and $F_{n+1}$ are not coprime, you take the smallest such $n$. Of course, $n$ cannot be $1$, since $F_1$ and $F_2$ are coprime. So, $n-1inmathbb N$ and it follows from what you proved that $F_{n-1}$ and $F_n$ are coprime too. This contradicts the definition of $n$.






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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

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            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            It is a waste of time to use modular arithmetic for this. What you proved was that$$kmid F_nwedge kmid F_{n+1}implies kmid F_{n-1},$$which is indeed correct. But, after having done this, I think that your best option is to say that you are using the well-ordering principle: if there was some $ninmathbb N$ such that $F_n$ and $F_{n+1}$ are not coprime, you take the smallest such $n$. Of course, $n$ cannot be $1$, since $F_1$ and $F_2$ are coprime. So, $n-1inmathbb N$ and it follows from what you proved that $F_{n-1}$ and $F_n$ are coprime too. This contradicts the definition of $n$.






            share|cite|improve this answer











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              3












              $begingroup$

              It is a waste of time to use modular arithmetic for this. What you proved was that$$kmid F_nwedge kmid F_{n+1}implies kmid F_{n-1},$$which is indeed correct. But, after having done this, I think that your best option is to say that you are using the well-ordering principle: if there was some $ninmathbb N$ such that $F_n$ and $F_{n+1}$ are not coprime, you take the smallest such $n$. Of course, $n$ cannot be $1$, since $F_1$ and $F_2$ are coprime. So, $n-1inmathbb N$ and it follows from what you proved that $F_{n-1}$ and $F_n$ are coprime too. This contradicts the definition of $n$.






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                It is a waste of time to use modular arithmetic for this. What you proved was that$$kmid F_nwedge kmid F_{n+1}implies kmid F_{n-1},$$which is indeed correct. But, after having done this, I think that your best option is to say that you are using the well-ordering principle: if there was some $ninmathbb N$ such that $F_n$ and $F_{n+1}$ are not coprime, you take the smallest such $n$. Of course, $n$ cannot be $1$, since $F_1$ and $F_2$ are coprime. So, $n-1inmathbb N$ and it follows from what you proved that $F_{n-1}$ and $F_n$ are coprime too. This contradicts the definition of $n$.






                share|cite|improve this answer











                $endgroup$



                It is a waste of time to use modular arithmetic for this. What you proved was that$$kmid F_nwedge kmid F_{n+1}implies kmid F_{n-1},$$which is indeed correct. But, after having done this, I think that your best option is to say that you are using the well-ordering principle: if there was some $ninmathbb N$ such that $F_n$ and $F_{n+1}$ are not coprime, you take the smallest such $n$. Of course, $n$ cannot be $1$, since $F_1$ and $F_2$ are coprime. So, $n-1inmathbb N$ and it follows from what you proved that $F_{n-1}$ and $F_n$ are coprime too. This contradicts the definition of $n$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 4 at 12:29

























                answered Jan 3 at 22:49









                José Carlos SantosJosé Carlos Santos

                155k22124227




                155k22124227






























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