Prove $n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0












0












$begingroup$


The statement I'm trying to prove is:



$n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0



I eventually need to prove $(k + 1)^3 + 7(k + 1) + 3$ is divisible by 3.



I don't really understand how to deal with $k + 1$, so I'm a little lost.



I've know that the base case of P(0) is true, but I'm not sure about proving the inductive case.






discrete-mathematics proof-verification induction







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$endgroup$












  • $begingroup$
    It's not true for $k=0$. It is true for $n=0$, but you didn't get to the variable $n$ until after saying $P(0)$ is true...
    $endgroup$
    – Thomas Andrews
    Feb 5 '16 at 17:45












  • $begingroup$
    It's only divisible by $3$ when $k+1$ is divisible by $3$, specifically.
    $endgroup$
    – Thomas Andrews
    Feb 5 '16 at 17:46








  • 2




    $begingroup$
    Maybe you meant $n^3 -7n+3$?
    $endgroup$
    – Lubin
    Feb 5 '16 at 18:03










  • $begingroup$
    yur statement is not true. peter.petrov has given a nice counter example.
    $endgroup$
    – Bhaskara-III
    Feb 5 '16 at 18:18
















0












$begingroup$


The statement I'm trying to prove is:



$n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0



I eventually need to prove $(k + 1)^3 + 7(k + 1) + 3$ is divisible by 3.



I don't really understand how to deal with $k + 1$, so I'm a little lost.



I've know that the base case of P(0) is true, but I'm not sure about proving the inductive case.






discrete-mathematics proof-verification induction







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's not true for $k=0$. It is true for $n=0$, but you didn't get to the variable $n$ until after saying $P(0)$ is true...
    $endgroup$
    – Thomas Andrews
    Feb 5 '16 at 17:45












  • $begingroup$
    It's only divisible by $3$ when $k+1$ is divisible by $3$, specifically.
    $endgroup$
    – Thomas Andrews
    Feb 5 '16 at 17:46








  • 2




    $begingroup$
    Maybe you meant $n^3 -7n+3$?
    $endgroup$
    – Lubin
    Feb 5 '16 at 18:03










  • $begingroup$
    yur statement is not true. peter.petrov has given a nice counter example.
    $endgroup$
    – Bhaskara-III
    Feb 5 '16 at 18:18














0












0








0


1



$begingroup$


The statement I'm trying to prove is:



$n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0



I eventually need to prove $(k + 1)^3 + 7(k + 1) + 3$ is divisible by 3.



I don't really understand how to deal with $k + 1$, so I'm a little lost.



I've know that the base case of P(0) is true, but I'm not sure about proving the inductive case.






discrete-mathematics proof-verification induction







share|cite|improve this question











$endgroup$




The statement I'm trying to prove is:



$n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0



I eventually need to prove $(k + 1)^3 + 7(k + 1) + 3$ is divisible by 3.



I don't really understand how to deal with $k + 1$, so I'm a little lost.



I've know that the base case of P(0) is true, but I'm not sure about proving the inductive case.






discrete-mathematics proof-verification induction




discrete-mathematics proof-verification induction






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 5 '16 at 17:53







123

















asked Feb 5 '16 at 17:41









123123

4551926




4551926












  • $begingroup$
    It's not true for $k=0$. It is true for $n=0$, but you didn't get to the variable $n$ until after saying $P(0)$ is true...
    $endgroup$
    – Thomas Andrews
    Feb 5 '16 at 17:45












  • $begingroup$
    It's only divisible by $3$ when $k+1$ is divisible by $3$, specifically.
    $endgroup$
    – Thomas Andrews
    Feb 5 '16 at 17:46








  • 2




    $begingroup$
    Maybe you meant $n^3 -7n+3$?
    $endgroup$
    – Lubin
    Feb 5 '16 at 18:03










  • $begingroup$
    yur statement is not true. peter.petrov has given a nice counter example.
    $endgroup$
    – Bhaskara-III
    Feb 5 '16 at 18:18


















  • $begingroup$
    It's not true for $k=0$. It is true for $n=0$, but you didn't get to the variable $n$ until after saying $P(0)$ is true...
    $endgroup$
    – Thomas Andrews
    Feb 5 '16 at 17:45












  • $begingroup$
    It's only divisible by $3$ when $k+1$ is divisible by $3$, specifically.
    $endgroup$
    – Thomas Andrews
    Feb 5 '16 at 17:46








  • 2




    $begingroup$
    Maybe you meant $n^3 -7n+3$?
    $endgroup$
    – Lubin
    Feb 5 '16 at 18:03










  • $begingroup$
    yur statement is not true. peter.petrov has given a nice counter example.
    $endgroup$
    – Bhaskara-III
    Feb 5 '16 at 18:18
















$begingroup$
It's not true for $k=0$. It is true for $n=0$, but you didn't get to the variable $n$ until after saying $P(0)$ is true...
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:45






$begingroup$
It's not true for $k=0$. It is true for $n=0$, but you didn't get to the variable $n$ until after saying $P(0)$ is true...
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:45














$begingroup$
It's only divisible by $3$ when $k+1$ is divisible by $3$, specifically.
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:46






$begingroup$
It's only divisible by $3$ when $k+1$ is divisible by $3$, specifically.
$endgroup$
– Thomas Andrews
Feb 5 '16 at 17:46






2




2




$begingroup$
Maybe you meant $n^3 -7n+3$?
$endgroup$
– Lubin
Feb 5 '16 at 18:03




$begingroup$
Maybe you meant $n^3 -7n+3$?
$endgroup$
– Lubin
Feb 5 '16 at 18:03












$begingroup$
yur statement is not true. peter.petrov has given a nice counter example.
$endgroup$
– Bhaskara-III
Feb 5 '16 at 18:18




$begingroup$
yur statement is not true. peter.petrov has given a nice counter example.
$endgroup$
– Bhaskara-III
Feb 5 '16 at 18:18










4 Answers
4






active

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4












$begingroup$

The statement is not true. Take $n=1$ as a counter example.



Since it's not true, you won't manage to prove it (by induction or otherwise).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    nice observation, +1
    $endgroup$
    – Bhaskara-III
    Feb 5 '16 at 18:17



















1












$begingroup$

Take $k+1=j$ and take modulo $3$. So the question is if $j^3+j$ is a multiple of $3$, but $j^3+j=j(j^2+1)$, which is never a multiple of $3$ if $j$ isn't.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Set $n=k+1$ then your expression is equivalent modulo 3 to
    $$
    n^3+7n+3 equiv n+7n+3 equiv 8n equiv -n pmod{3}.
    $$
    In particular, this is a multiple of $3$ iff $3$ divides $k+1$.






    share|cite|improve this answer









    $endgroup$





















      -1












      $begingroup$

      The given statement $n^3 + 7n + 3$ is divisible by 3 for $n ≥ 0$ is not true, as detailed below
      P(n): n^3. + 7n + 3 is divisible by 3
      P(1): 1^3 + 7(1) + 3 = 11 which is not divisible by 3
      P(2): 2^3 + 7(2) + 3 = 25 which is also not divisible by 3
      Hence given statement is not true.



      I presume that statement may be n^3 - 7n + 3 is divisible by 3 for n ≥ 0
      In that case Proof is detailed below.



      P(n): n^3 - 7n + 3 is divisible by 3
      P(1): (1)^3 -7(1) + 3 = 1 -7 + 3 = -3 which is divisible by 3



      Assume P(k) is true for some positive integer k, i.e
      P(k): (k)^3 - 7k + 3 is divisible by 3
      We can write (k)^3 - 7k + 3 = 3d, where d ∈ N ………….. (1)
      We shall prove that P(k+1) is also true, whenever P(k) is true. Now we have
      P(k+1): (k+1)^3 - 7(k+1) + 3

      = k^3 + 1^3 +3.k.1(k+1) - 7(k+1) + 3 (Using (a+b)^3 = a^3 + b^3 + 3ab(a+b) 



          = k^3 + 1 +3k(k+1) - 7(k+1) + 3  
      
    = k^3 + 1 + 3k^2 + 3k -7k - 7 +3
      
    = k^3 - 7k + 3 + 3k^2 +3k - 6  (by rearranging)
      
    = 3d + 3(k^2 + k - 2 ).........{From (1)}      
      
    = 3d + 3(k^2 + 2k - k - 2 ) 
      
    = 3d + 3 {k(k +2)-(k+2)}
      
    = 3d + 3 (k-1)(k+2)
      
    = 3 {d +(k-1)(k+2)}
      
    = 3 q,  where q = {d +(k-1)(k+2)} which is some positive integer


      From the last line, we see (k+1)^3 - 7(k+1) + 3 ) is divisible by 3.
      Thus P(k+1) is true, whenever P(k) is true.
      Hence, by the PMI, the given statement is true for all natural numbers n ≥ 0






      share|cite|improve this answer











      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        The statement is not true. Take $n=1$ as a counter example.



        Since it's not true, you won't manage to prove it (by induction or otherwise).






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          nice observation, +1
          $endgroup$
          – Bhaskara-III
          Feb 5 '16 at 18:17
















        4












        $begingroup$

        The statement is not true. Take $n=1$ as a counter example.



        Since it's not true, you won't manage to prove it (by induction or otherwise).






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          nice observation, +1
          $endgroup$
          – Bhaskara-III
          Feb 5 '16 at 18:17














        4












        4








        4





        $begingroup$

        The statement is not true. Take $n=1$ as a counter example.



        Since it's not true, you won't manage to prove it (by induction or otherwise).






        share|cite|improve this answer









        $endgroup$



        The statement is not true. Take $n=1$ as a counter example.



        Since it's not true, you won't manage to prove it (by induction or otherwise).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 5 '16 at 17:42









        peter.petrovpeter.petrov

        5,441821




        5,441821












        • $begingroup$
          nice observation, +1
          $endgroup$
          – Bhaskara-III
          Feb 5 '16 at 18:17


















        • $begingroup$
          nice observation, +1
          $endgroup$
          – Bhaskara-III
          Feb 5 '16 at 18:17
















        $begingroup$
        nice observation, +1
        $endgroup$
        – Bhaskara-III
        Feb 5 '16 at 18:17




        $begingroup$
        nice observation, +1
        $endgroup$
        – Bhaskara-III
        Feb 5 '16 at 18:17











        1












        $begingroup$

        Take $k+1=j$ and take modulo $3$. So the question is if $j^3+j$ is a multiple of $3$, but $j^3+j=j(j^2+1)$, which is never a multiple of $3$ if $j$ isn't.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Take $k+1=j$ and take modulo $3$. So the question is if $j^3+j$ is a multiple of $3$, but $j^3+j=j(j^2+1)$, which is never a multiple of $3$ if $j$ isn't.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Take $k+1=j$ and take modulo $3$. So the question is if $j^3+j$ is a multiple of $3$, but $j^3+j=j(j^2+1)$, which is never a multiple of $3$ if $j$ isn't.






            share|cite|improve this answer









            $endgroup$



            Take $k+1=j$ and take modulo $3$. So the question is if $j^3+j$ is a multiple of $3$, but $j^3+j=j(j^2+1)$, which is never a multiple of $3$ if $j$ isn't.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 5 '16 at 17:46









            ajotatxeajotatxe

            54k24090




            54k24090























                1












                $begingroup$

                Set $n=k+1$ then your expression is equivalent modulo 3 to
                $$
                n^3+7n+3 equiv n+7n+3 equiv 8n equiv -n pmod{3}.
                $$
                In particular, this is a multiple of $3$ iff $3$ divides $k+1$.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Set $n=k+1$ then your expression is equivalent modulo 3 to
                  $$
                  n^3+7n+3 equiv n+7n+3 equiv 8n equiv -n pmod{3}.
                  $$
                  In particular, this is a multiple of $3$ iff $3$ divides $k+1$.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Set $n=k+1$ then your expression is equivalent modulo 3 to
                    $$
                    n^3+7n+3 equiv n+7n+3 equiv 8n equiv -n pmod{3}.
                    $$
                    In particular, this is a multiple of $3$ iff $3$ divides $k+1$.






                    share|cite|improve this answer









                    $endgroup$



                    Set $n=k+1$ then your expression is equivalent modulo 3 to
                    $$
                    n^3+7n+3 equiv n+7n+3 equiv 8n equiv -n pmod{3}.
                    $$
                    In particular, this is a multiple of $3$ iff $3$ divides $k+1$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Feb 5 '16 at 17:47









                    Paolo LeonettiPaolo Leonetti

                    11.5k21550




                    11.5k21550























                        -1












                        $begingroup$

                        The given statement $n^3 + 7n + 3$ is divisible by 3 for $n ≥ 0$ is not true, as detailed below
                        P(n): n^3. + 7n + 3 is divisible by 3
                        P(1): 1^3 + 7(1) + 3 = 11 which is not divisible by 3
                        P(2): 2^3 + 7(2) + 3 = 25 which is also not divisible by 3
                        Hence given statement is not true.



                        I presume that statement may be n^3 - 7n + 3 is divisible by 3 for n ≥ 0
                        In that case Proof is detailed below.



                        P(n): n^3 - 7n + 3 is divisible by 3
                        P(1): (1)^3 -7(1) + 3 = 1 -7 + 3 = -3 which is divisible by 3



                        Assume P(k) is true for some positive integer k, i.e
                        P(k): (k)^3 - 7k + 3 is divisible by 3
                        We can write (k)^3 - 7k + 3 = 3d, where d ∈ N ………….. (1)
                        We shall prove that P(k+1) is also true, whenever P(k) is true. Now we have
                        P(k+1): (k+1)^3 - 7(k+1) + 3

                        = k^3 + 1^3 +3.k.1(k+1) - 7(k+1) + 3 (Using (a+b)^3 = a^3 + b^3 + 3ab(a+b) 



                            = k^3 + 1 +3k(k+1) - 7(k+1) + 3  
                        
    = k^3 + 1 + 3k^2 + 3k -7k - 7 +3
                        
    = k^3 - 7k + 3 + 3k^2 +3k - 6  (by rearranging)
                        
    = 3d + 3(k^2 + k - 2 ).........{From (1)}      
                        
    = 3d + 3(k^2 + 2k - k - 2 ) 
                        
    = 3d + 3 {k(k +2)-(k+2)}
                        
    = 3d + 3 (k-1)(k+2)
                        
    = 3 {d +(k-1)(k+2)}
                        
    = 3 q,  where q = {d +(k-1)(k+2)} which is some positive integer


                        From the last line, we see (k+1)^3 - 7(k+1) + 3 ) is divisible by 3.
                        Thus P(k+1) is true, whenever P(k) is true.
                        Hence, by the PMI, the given statement is true for all natural numbers n ≥ 0






                        share|cite|improve this answer











                        $endgroup$


















                          -1












                          $begingroup$

                          The given statement $n^3 + 7n + 3$ is divisible by 3 for $n ≥ 0$ is not true, as detailed below
                          P(n): n^3. + 7n + 3 is divisible by 3
                          P(1): 1^3 + 7(1) + 3 = 11 which is not divisible by 3
                          P(2): 2^3 + 7(2) + 3 = 25 which is also not divisible by 3
                          Hence given statement is not true.



                          I presume that statement may be n^3 - 7n + 3 is divisible by 3 for n ≥ 0
                          In that case Proof is detailed below.



                          P(n): n^3 - 7n + 3 is divisible by 3
                          P(1): (1)^3 -7(1) + 3 = 1 -7 + 3 = -3 which is divisible by 3



                          Assume P(k) is true for some positive integer k, i.e
                          P(k): (k)^3 - 7k + 3 is divisible by 3
                          We can write (k)^3 - 7k + 3 = 3d, where d ∈ N ………….. (1)
                          We shall prove that P(k+1) is also true, whenever P(k) is true. Now we have
                          P(k+1): (k+1)^3 - 7(k+1) + 3

                          = k^3 + 1^3 +3.k.1(k+1) - 7(k+1) + 3 (Using (a+b)^3 = a^3 + b^3 + 3ab(a+b) 



                              = k^3 + 1 +3k(k+1) - 7(k+1) + 3  
                          
    = k^3 + 1 + 3k^2 + 3k -7k - 7 +3
                          
    = k^3 - 7k + 3 + 3k^2 +3k - 6  (by rearranging)
                          
    = 3d + 3(k^2 + k - 2 ).........{From (1)}      
                          
    = 3d + 3(k^2 + 2k - k - 2 ) 
                          
    = 3d + 3 {k(k +2)-(k+2)}
                          
    = 3d + 3 (k-1)(k+2)
                          
    = 3 {d +(k-1)(k+2)}
                          
    = 3 q,  where q = {d +(k-1)(k+2)} which is some positive integer


                          From the last line, we see (k+1)^3 - 7(k+1) + 3 ) is divisible by 3.
                          Thus P(k+1) is true, whenever P(k) is true.
                          Hence, by the PMI, the given statement is true for all natural numbers n ≥ 0






                          share|cite|improve this answer











                          $endgroup$
















                            -1












                            -1








                            -1





                            $begingroup$

                            The given statement $n^3 + 7n + 3$ is divisible by 3 for $n ≥ 0$ is not true, as detailed below
                            P(n): n^3. + 7n + 3 is divisible by 3
                            P(1): 1^3 + 7(1) + 3 = 11 which is not divisible by 3
                            P(2): 2^3 + 7(2) + 3 = 25 which is also not divisible by 3
                            Hence given statement is not true.



                            I presume that statement may be n^3 - 7n + 3 is divisible by 3 for n ≥ 0
                            In that case Proof is detailed below.



                            P(n): n^3 - 7n + 3 is divisible by 3
                            P(1): (1)^3 -7(1) + 3 = 1 -7 + 3 = -3 which is divisible by 3



                            Assume P(k) is true for some positive integer k, i.e
                            P(k): (k)^3 - 7k + 3 is divisible by 3
                            We can write (k)^3 - 7k + 3 = 3d, where d ∈ N ………….. (1)
                            We shall prove that P(k+1) is also true, whenever P(k) is true. Now we have
                            P(k+1): (k+1)^3 - 7(k+1) + 3

                            = k^3 + 1^3 +3.k.1(k+1) - 7(k+1) + 3 (Using (a+b)^3 = a^3 + b^3 + 3ab(a+b) 



                                = k^3 + 1 +3k(k+1) - 7(k+1) + 3  
                            
    = k^3 + 1 + 3k^2 + 3k -7k - 7 +3
                            
    = k^3 - 7k + 3 + 3k^2 +3k - 6  (by rearranging)
                            
    = 3d + 3(k^2 + k - 2 ).........{From (1)}      
                            
    = 3d + 3(k^2 + 2k - k - 2 ) 
                            
    = 3d + 3 {k(k +2)-(k+2)}
                            
    = 3d + 3 (k-1)(k+2)
                            
    = 3 {d +(k-1)(k+2)}
                            
    = 3 q,  where q = {d +(k-1)(k+2)} which is some positive integer


                            From the last line, we see (k+1)^3 - 7(k+1) + 3 ) is divisible by 3.
                            Thus P(k+1) is true, whenever P(k) is true.
                            Hence, by the PMI, the given statement is true for all natural numbers n ≥ 0






                            share|cite|improve this answer











                            $endgroup$



                            The given statement $n^3 + 7n + 3$ is divisible by 3 for $n ≥ 0$ is not true, as detailed below
                            P(n): n^3. + 7n + 3 is divisible by 3
                            P(1): 1^3 + 7(1) + 3 = 11 which is not divisible by 3
                            P(2): 2^3 + 7(2) + 3 = 25 which is also not divisible by 3
                            Hence given statement is not true.



                            I presume that statement may be n^3 - 7n + 3 is divisible by 3 for n ≥ 0
                            In that case Proof is detailed below.



                            P(n): n^3 - 7n + 3 is divisible by 3
                            P(1): (1)^3 -7(1) + 3 = 1 -7 + 3 = -3 which is divisible by 3



                            Assume P(k) is true for some positive integer k, i.e
                            P(k): (k)^3 - 7k + 3 is divisible by 3
                            We can write (k)^3 - 7k + 3 = 3d, where d ∈ N ………….. (1)
                            We shall prove that P(k+1) is also true, whenever P(k) is true. Now we have
                            P(k+1): (k+1)^3 - 7(k+1) + 3

                            = k^3 + 1^3 +3.k.1(k+1) - 7(k+1) + 3 (Using (a+b)^3 = a^3 + b^3 + 3ab(a+b) 



                                = k^3 + 1 +3k(k+1) - 7(k+1) + 3  
                            
    = k^3 + 1 + 3k^2 + 3k -7k - 7 +3
                            
    = k^3 - 7k + 3 + 3k^2 +3k - 6  (by rearranging)
                            
    = 3d + 3(k^2 + k - 2 ).........{From (1)}      
                            
    = 3d + 3(k^2 + 2k - k - 2 ) 
                            
    = 3d + 3 {k(k +2)-(k+2)}
                            
    = 3d + 3 (k-1)(k+2)
                            
    = 3 {d +(k-1)(k+2)}
                            
    = 3 q,  where q = {d +(k-1)(k+2)} which is some positive integer


                            From the last line, we see (k+1)^3 - 7(k+1) + 3 ) is divisible by 3.
                            Thus P(k+1) is true, whenever P(k) is true.
                            Hence, by the PMI, the given statement is true for all natural numbers n ≥ 0







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                            edited Dec 28 '18 at 21:08









                            Namaste

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                            answered Dec 28 '18 at 3:09









                            Nammalwar ThiruvengadamNammalwar Thiruvengadam

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