Question pertaining to perfect powers in sums of consecutive integers












1












$begingroup$


I came across this question in the book Challenge and Thrill of Pre-college mathematics:




Prove that there are infinitely many sets of five consecutive integers $a,b,c,d,e$ such that $a+b+c+d+e$ is a perfect cube, and $b+c+d$ is a perfect square.




I took the numbers to be $(n-2),(n-1),n,(n+1),(n+2)$, and then it was not very difficult to find that what was required to prove was the existence of infinite numbers $n$ such that $5n$ and $3n$ are perfect cubes and squares respectively.



After playing around with prime factorizations for a while, I discovered that $675$ was such a number, followed by $3^9cdot5^2$. Clearly this has something to do with the prime factorization of these integers.



Futhermore, I think it is correct to assume that that multiplying $675$ with any number $p^k$ where $p$ is a prime and where $kmid3$ and $kmid2$, such as multiplying by $2^{6}$, which gives us $43200$, which is indeed such a number.



My problem is that I have no idea how I am supposed to write such a proof formally, or even if this is a proof. While I do believe that I have generated an infinite number of positive integers which obey the question, how do I explain it? What am I missing?






elementary-number-theory proof-writing







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    My below answer is a different way to show that infinite many solutions exist, but your argument is perfectly right. You have found one solution and showed that we can multiply it how often we want with $64$ to get another solution. This clearly shows that infinite many solutions must exist.
    $endgroup$
    – Peter
    Jan 19 at 15:09












  • $begingroup$
    Do you think you can extend this result to more primes?
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 19 at 15:17






  • 1




    $begingroup$
    Extend the $p^k$ result or to some integer $n$ where $pn$ and $qn$ are powers for prime $p$ and $q$?
    $endgroup$
    – Naman Kumar
    Jan 19 at 15:19
















1












$begingroup$


I came across this question in the book Challenge and Thrill of Pre-college mathematics:




Prove that there are infinitely many sets of five consecutive integers $a,b,c,d,e$ such that $a+b+c+d+e$ is a perfect cube, and $b+c+d$ is a perfect square.




I took the numbers to be $(n-2),(n-1),n,(n+1),(n+2)$, and then it was not very difficult to find that what was required to prove was the existence of infinite numbers $n$ such that $5n$ and $3n$ are perfect cubes and squares respectively.



After playing around with prime factorizations for a while, I discovered that $675$ was such a number, followed by $3^9cdot5^2$. Clearly this has something to do with the prime factorization of these integers.



Futhermore, I think it is correct to assume that that multiplying $675$ with any number $p^k$ where $p$ is a prime and where $kmid3$ and $kmid2$, such as multiplying by $2^{6}$, which gives us $43200$, which is indeed such a number.



My problem is that I have no idea how I am supposed to write such a proof formally, or even if this is a proof. While I do believe that I have generated an infinite number of positive integers which obey the question, how do I explain it? What am I missing?






elementary-number-theory proof-writing







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    My below answer is a different way to show that infinite many solutions exist, but your argument is perfectly right. You have found one solution and showed that we can multiply it how often we want with $64$ to get another solution. This clearly shows that infinite many solutions must exist.
    $endgroup$
    – Peter
    Jan 19 at 15:09












  • $begingroup$
    Do you think you can extend this result to more primes?
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 19 at 15:17






  • 1




    $begingroup$
    Extend the $p^k$ result or to some integer $n$ where $pn$ and $qn$ are powers for prime $p$ and $q$?
    $endgroup$
    – Naman Kumar
    Jan 19 at 15:19














1












1








1





$begingroup$


I came across this question in the book Challenge and Thrill of Pre-college mathematics:




Prove that there are infinitely many sets of five consecutive integers $a,b,c,d,e$ such that $a+b+c+d+e$ is a perfect cube, and $b+c+d$ is a perfect square.




I took the numbers to be $(n-2),(n-1),n,(n+1),(n+2)$, and then it was not very difficult to find that what was required to prove was the existence of infinite numbers $n$ such that $5n$ and $3n$ are perfect cubes and squares respectively.



After playing around with prime factorizations for a while, I discovered that $675$ was such a number, followed by $3^9cdot5^2$. Clearly this has something to do with the prime factorization of these integers.



Futhermore, I think it is correct to assume that that multiplying $675$ with any number $p^k$ where $p$ is a prime and where $kmid3$ and $kmid2$, such as multiplying by $2^{6}$, which gives us $43200$, which is indeed such a number.



My problem is that I have no idea how I am supposed to write such a proof formally, or even if this is a proof. While I do believe that I have generated an infinite number of positive integers which obey the question, how do I explain it? What am I missing?






elementary-number-theory proof-writing







share|cite|improve this question









$endgroup$




I came across this question in the book Challenge and Thrill of Pre-college mathematics:




Prove that there are infinitely many sets of five consecutive integers $a,b,c,d,e$ such that $a+b+c+d+e$ is a perfect cube, and $b+c+d$ is a perfect square.




I took the numbers to be $(n-2),(n-1),n,(n+1),(n+2)$, and then it was not very difficult to find that what was required to prove was the existence of infinite numbers $n$ such that $5n$ and $3n$ are perfect cubes and squares respectively.



After playing around with prime factorizations for a while, I discovered that $675$ was such a number, followed by $3^9cdot5^2$. Clearly this has something to do with the prime factorization of these integers.



Futhermore, I think it is correct to assume that that multiplying $675$ with any number $p^k$ where $p$ is a prime and where $kmid3$ and $kmid2$, such as multiplying by $2^{6}$, which gives us $43200$, which is indeed such a number.



My problem is that I have no idea how I am supposed to write such a proof formally, or even if this is a proof. While I do believe that I have generated an infinite number of positive integers which obey the question, how do I explain it? What am I missing?






elementary-number-theory proof-writing




elementary-number-theory proof-writing






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 19 at 14:55









Naman KumarNaman Kumar

23813




23813








  • 2




    $begingroup$
    My below answer is a different way to show that infinite many solutions exist, but your argument is perfectly right. You have found one solution and showed that we can multiply it how often we want with $64$ to get another solution. This clearly shows that infinite many solutions must exist.
    $endgroup$
    – Peter
    Jan 19 at 15:09












  • $begingroup$
    Do you think you can extend this result to more primes?
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 19 at 15:17






  • 1




    $begingroup$
    Extend the $p^k$ result or to some integer $n$ where $pn$ and $qn$ are powers for prime $p$ and $q$?
    $endgroup$
    – Naman Kumar
    Jan 19 at 15:19














  • 2




    $begingroup$
    My below answer is a different way to show that infinite many solutions exist, but your argument is perfectly right. You have found one solution and showed that we can multiply it how often we want with $64$ to get another solution. This clearly shows that infinite many solutions must exist.
    $endgroup$
    – Peter
    Jan 19 at 15:09












  • $begingroup$
    Do you think you can extend this result to more primes?
    $endgroup$
    – астон вілла олоф мэллбэрг
    Jan 19 at 15:17






  • 1




    $begingroup$
    Extend the $p^k$ result or to some integer $n$ where $pn$ and $qn$ are powers for prime $p$ and $q$?
    $endgroup$
    – Naman Kumar
    Jan 19 at 15:19








2




2




$begingroup$
My below answer is a different way to show that infinite many solutions exist, but your argument is perfectly right. You have found one solution and showed that we can multiply it how often we want with $64$ to get another solution. This clearly shows that infinite many solutions must exist.
$endgroup$
– Peter
Jan 19 at 15:09






$begingroup$
My below answer is a different way to show that infinite many solutions exist, but your argument is perfectly right. You have found one solution and showed that we can multiply it how often we want with $64$ to get another solution. This clearly shows that infinite many solutions must exist.
$endgroup$
– Peter
Jan 19 at 15:09














$begingroup$
Do you think you can extend this result to more primes?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 15:17




$begingroup$
Do you think you can extend this result to more primes?
$endgroup$
– астон вілла олоф мэллбэрг
Jan 19 at 15:17




1




1




$begingroup$
Extend the $p^k$ result or to some integer $n$ where $pn$ and $qn$ are powers for prime $p$ and $q$?
$endgroup$
– Naman Kumar
Jan 19 at 15:19




$begingroup$
Extend the $p^k$ result or to some integer $n$ where $pn$ and $qn$ are powers for prime $p$ and $q$?
$endgroup$
– Naman Kumar
Jan 19 at 15:19










3 Answers
3






active

oldest

votes


















1












$begingroup$

You have to find positive integers $nge 3$, such that $3n$ is a perfect square and $5n$ is a perfect cube.



Consider $$n=3^acdot 5^b$$ with positive integers $a$ and $b$. $3n$ is a perfect square, if $a$ is odd and $b$ is even. $5n$ is a perfect cube, if $a$ is divisible by $3$ and $b$ has the form $3k+2$. Hence, it is enough that $a$ is of the form $6s+3$ and $b$ of the form $6t+2$ , giving infinite many solutions.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    You already have it all for demonstrating there are infinite such numbers.



    $$n=675cdot p^6$$
    for primes $p>5$ already gives you infinitely many solutions.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      You basically want infinitely many $n$ with $5n$ a perfect cube, and $3n$ a perfect square. Indeed, your last argument is enough to produce infinitely such $n$, since $n = 675p^{6l}$, for any prime(in fact any number) $p$ and any $l$ satisfies the given conditions. This is easy to see because :
      $$
      5n = 5 times 675 p^{6l} = 5^33^3p^{6l} = (15p^{2l})^3 quad ; quad 3n = 3 times 675 p^{6l} = 5^23^4p^{6l} = (45p^{3l})^2
      $$



      Since both $p$ and $l$ may infinitely vary, we see (from prime factorization) that we can produce infinitely many values of $n$, and hence infinitely many tuples $(a,b,c,d,e) = (n-2,n-1,n,n+1,n+2)$.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        You have to find positive integers $nge 3$, such that $3n$ is a perfect square and $5n$ is a perfect cube.



        Consider $$n=3^acdot 5^b$$ with positive integers $a$ and $b$. $3n$ is a perfect square, if $a$ is odd and $b$ is even. $5n$ is a perfect cube, if $a$ is divisible by $3$ and $b$ has the form $3k+2$. Hence, it is enough that $a$ is of the form $6s+3$ and $b$ of the form $6t+2$ , giving infinite many solutions.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          You have to find positive integers $nge 3$, such that $3n$ is a perfect square and $5n$ is a perfect cube.



          Consider $$n=3^acdot 5^b$$ with positive integers $a$ and $b$. $3n$ is a perfect square, if $a$ is odd and $b$ is even. $5n$ is a perfect cube, if $a$ is divisible by $3$ and $b$ has the form $3k+2$. Hence, it is enough that $a$ is of the form $6s+3$ and $b$ of the form $6t+2$ , giving infinite many solutions.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            You have to find positive integers $nge 3$, such that $3n$ is a perfect square and $5n$ is a perfect cube.



            Consider $$n=3^acdot 5^b$$ with positive integers $a$ and $b$. $3n$ is a perfect square, if $a$ is odd and $b$ is even. $5n$ is a perfect cube, if $a$ is divisible by $3$ and $b$ has the form $3k+2$. Hence, it is enough that $a$ is of the form $6s+3$ and $b$ of the form $6t+2$ , giving infinite many solutions.






            share|cite|improve this answer









            $endgroup$



            You have to find positive integers $nge 3$, such that $3n$ is a perfect square and $5n$ is a perfect cube.



            Consider $$n=3^acdot 5^b$$ with positive integers $a$ and $b$. $3n$ is a perfect square, if $a$ is odd and $b$ is even. $5n$ is a perfect cube, if $a$ is divisible by $3$ and $b$ has the form $3k+2$. Hence, it is enough that $a$ is of the form $6s+3$ and $b$ of the form $6t+2$ , giving infinite many solutions.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 19 at 15:03









            PeterPeter

            47.6k1039131




            47.6k1039131























                1












                $begingroup$

                You already have it all for demonstrating there are infinite such numbers.



                $$n=675cdot p^6$$
                for primes $p>5$ already gives you infinitely many solutions.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  You already have it all for demonstrating there are infinite such numbers.



                  $$n=675cdot p^6$$
                  for primes $p>5$ already gives you infinitely many solutions.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    You already have it all for demonstrating there are infinite such numbers.



                    $$n=675cdot p^6$$
                    for primes $p>5$ already gives you infinitely many solutions.






                    share|cite|improve this answer









                    $endgroup$



                    You already have it all for demonstrating there are infinite such numbers.



                    $$n=675cdot p^6$$
                    for primes $p>5$ already gives you infinitely many solutions.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 19 at 15:05









                    BerciBerci

                    60.9k23673




                    60.9k23673























                        1












                        $begingroup$

                        You basically want infinitely many $n$ with $5n$ a perfect cube, and $3n$ a perfect square. Indeed, your last argument is enough to produce infinitely such $n$, since $n = 675p^{6l}$, for any prime(in fact any number) $p$ and any $l$ satisfies the given conditions. This is easy to see because :
                        $$
                        5n = 5 times 675 p^{6l} = 5^33^3p^{6l} = (15p^{2l})^3 quad ; quad 3n = 3 times 675 p^{6l} = 5^23^4p^{6l} = (45p^{3l})^2
                        $$



                        Since both $p$ and $l$ may infinitely vary, we see (from prime factorization) that we can produce infinitely many values of $n$, and hence infinitely many tuples $(a,b,c,d,e) = (n-2,n-1,n,n+1,n+2)$.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          You basically want infinitely many $n$ with $5n$ a perfect cube, and $3n$ a perfect square. Indeed, your last argument is enough to produce infinitely such $n$, since $n = 675p^{6l}$, for any prime(in fact any number) $p$ and any $l$ satisfies the given conditions. This is easy to see because :
                          $$
                          5n = 5 times 675 p^{6l} = 5^33^3p^{6l} = (15p^{2l})^3 quad ; quad 3n = 3 times 675 p^{6l} = 5^23^4p^{6l} = (45p^{3l})^2
                          $$



                          Since both $p$ and $l$ may infinitely vary, we see (from prime factorization) that we can produce infinitely many values of $n$, and hence infinitely many tuples $(a,b,c,d,e) = (n-2,n-1,n,n+1,n+2)$.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            You basically want infinitely many $n$ with $5n$ a perfect cube, and $3n$ a perfect square. Indeed, your last argument is enough to produce infinitely such $n$, since $n = 675p^{6l}$, for any prime(in fact any number) $p$ and any $l$ satisfies the given conditions. This is easy to see because :
                            $$
                            5n = 5 times 675 p^{6l} = 5^33^3p^{6l} = (15p^{2l})^3 quad ; quad 3n = 3 times 675 p^{6l} = 5^23^4p^{6l} = (45p^{3l})^2
                            $$



                            Since both $p$ and $l$ may infinitely vary, we see (from prime factorization) that we can produce infinitely many values of $n$, and hence infinitely many tuples $(a,b,c,d,e) = (n-2,n-1,n,n+1,n+2)$.






                            share|cite|improve this answer









                            $endgroup$



                            You basically want infinitely many $n$ with $5n$ a perfect cube, and $3n$ a perfect square. Indeed, your last argument is enough to produce infinitely such $n$, since $n = 675p^{6l}$, for any prime(in fact any number) $p$ and any $l$ satisfies the given conditions. This is easy to see because :
                            $$
                            5n = 5 times 675 p^{6l} = 5^33^3p^{6l} = (15p^{2l})^3 quad ; quad 3n = 3 times 675 p^{6l} = 5^23^4p^{6l} = (45p^{3l})^2
                            $$



                            Since both $p$ and $l$ may infinitely vary, we see (from prime factorization) that we can produce infinitely many values of $n$, and hence infinitely many tuples $(a,b,c,d,e) = (n-2,n-1,n,n+1,n+2)$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 19 at 15:06









                            астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

                            38.5k33376




                            38.5k33376






























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