Prove that for all positive integers $n>1, (n^2+1)-n$ is not a perfect square.












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In this hard question, I tried to use mathematical induction but couldn't as it is a hard problem. I have been trying for a week now so I came to stack exchange with this question. Any help would be nice.



Also the problems goes on and asks to prove for $2(n^2+1)-n$






elementary-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you really mean to write $1(n^2+1)-n$ or is there a typo?
    $endgroup$
    – bof
    Jan 5 at 7:41










  • $begingroup$
    I think it is supposed to be all positive integers greater than $1$, as when $n=1$ it is a square.
    $endgroup$
    – Erik Parkinson
    Jan 5 at 7:43










  • $begingroup$
    I forgot to add the second part
    $endgroup$
    – user587054
    Jan 5 at 8:19
















0












$begingroup$


In this hard question, I tried to use mathematical induction but couldn't as it is a hard problem. I have been trying for a week now so I came to stack exchange with this question. Any help would be nice.



Also the problems goes on and asks to prove for $2(n^2+1)-n$






elementary-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you really mean to write $1(n^2+1)-n$ or is there a typo?
    $endgroup$
    – bof
    Jan 5 at 7:41










  • $begingroup$
    I think it is supposed to be all positive integers greater than $1$, as when $n=1$ it is a square.
    $endgroup$
    – Erik Parkinson
    Jan 5 at 7:43










  • $begingroup$
    I forgot to add the second part
    $endgroup$
    – user587054
    Jan 5 at 8:19














0












0








0





$begingroup$


In this hard question, I tried to use mathematical induction but couldn't as it is a hard problem. I have been trying for a week now so I came to stack exchange with this question. Any help would be nice.



Also the problems goes on and asks to prove for $2(n^2+1)-n$






elementary-number-theory







share|cite|improve this question











$endgroup$




In this hard question, I tried to use mathematical induction but couldn't as it is a hard problem. I have been trying for a week now so I came to stack exchange with this question. Any help would be nice.



Also the problems goes on and asks to prove for $2(n^2+1)-n$






elementary-number-theory




elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 8:19







user587054

















asked Jan 5 at 7:36









user587054user587054

46511




46511












  • $begingroup$
    Did you really mean to write $1(n^2+1)-n$ or is there a typo?
    $endgroup$
    – bof
    Jan 5 at 7:41










  • $begingroup$
    I think it is supposed to be all positive integers greater than $1$, as when $n=1$ it is a square.
    $endgroup$
    – Erik Parkinson
    Jan 5 at 7:43










  • $begingroup$
    I forgot to add the second part
    $endgroup$
    – user587054
    Jan 5 at 8:19


















  • $begingroup$
    Did you really mean to write $1(n^2+1)-n$ or is there a typo?
    $endgroup$
    – bof
    Jan 5 at 7:41










  • $begingroup$
    I think it is supposed to be all positive integers greater than $1$, as when $n=1$ it is a square.
    $endgroup$
    – Erik Parkinson
    Jan 5 at 7:43










  • $begingroup$
    I forgot to add the second part
    $endgroup$
    – user587054
    Jan 5 at 8:19
















$begingroup$
Did you really mean to write $1(n^2+1)-n$ or is there a typo?
$endgroup$
– bof
Jan 5 at 7:41




$begingroup$
Did you really mean to write $1(n^2+1)-n$ or is there a typo?
$endgroup$
– bof
Jan 5 at 7:41












$begingroup$
I think it is supposed to be all positive integers greater than $1$, as when $n=1$ it is a square.
$endgroup$
– Erik Parkinson
Jan 5 at 7:43




$begingroup$
I think it is supposed to be all positive integers greater than $1$, as when $n=1$ it is a square.
$endgroup$
– Erik Parkinson
Jan 5 at 7:43












$begingroup$
I forgot to add the second part
$endgroup$
– user587054
Jan 5 at 8:19




$begingroup$
I forgot to add the second part
$endgroup$
– user587054
Jan 5 at 8:19










3 Answers
3






active

oldest

votes


















3












$begingroup$

The trick is that for all $n ge 2$,
$$(n-1)^2 = n^2-2n+1 < n^2-n+1 < n^2$$



So $n^2-n+1$ will lie between all consecutive square numbers and thus can't be square.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Note that for $n^2-n+1$ and $n >1$, the following inequality holds true:



    $$n^2-2n+1 < n^2-n+1 < n^2$$



    $$(n-1)^2 < n^2-n+1 < n^2$$



    where $(n-1)^2$ and $n^2$ are two consecutive squares. What can be concluded from this?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you but can you please help me with the second part I updated on the post. Thank you anyways
      $endgroup$
      – user587054
      Jan 5 at 8:39



















    0












    $begingroup$

    A hint for the second part. Consider the possibilities for $n pmod3$, that is, $n$ may be of the form $3k$, $3k+1$ or $3k+2$. Do any of these $n$ yield expressions for $2(n^2+1)-n$ that are possible squares $pmod3$? And if any is, is it also a possible square $pmod9$?






    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









      3












      $begingroup$

      The trick is that for all $n ge 2$,
      $$(n-1)^2 = n^2-2n+1 < n^2-n+1 < n^2$$



      So $n^2-n+1$ will lie between all consecutive square numbers and thus can't be square.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        The trick is that for all $n ge 2$,
        $$(n-1)^2 = n^2-2n+1 < n^2-n+1 < n^2$$



        So $n^2-n+1$ will lie between all consecutive square numbers and thus can't be square.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          The trick is that for all $n ge 2$,
          $$(n-1)^2 = n^2-2n+1 < n^2-n+1 < n^2$$



          So $n^2-n+1$ will lie between all consecutive square numbers and thus can't be square.






          share|cite|improve this answer









          $endgroup$



          The trick is that for all $n ge 2$,
          $$(n-1)^2 = n^2-2n+1 < n^2-n+1 < n^2$$



          So $n^2-n+1$ will lie between all consecutive square numbers and thus can't be square.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 5 at 7:42









          Erik ParkinsonErik Parkinson

          9159




          9159























              1












              $begingroup$

              Note that for $n^2-n+1$ and $n >1$, the following inequality holds true:



              $$n^2-2n+1 < n^2-n+1 < n^2$$



              $$(n-1)^2 < n^2-n+1 < n^2$$



              where $(n-1)^2$ and $n^2$ are two consecutive squares. What can be concluded from this?






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Thank you but can you please help me with the second part I updated on the post. Thank you anyways
                $endgroup$
                – user587054
                Jan 5 at 8:39
















              1












              $begingroup$

              Note that for $n^2-n+1$ and $n >1$, the following inequality holds true:



              $$n^2-2n+1 < n^2-n+1 < n^2$$



              $$(n-1)^2 < n^2-n+1 < n^2$$



              where $(n-1)^2$ and $n^2$ are two consecutive squares. What can be concluded from this?






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Thank you but can you please help me with the second part I updated on the post. Thank you anyways
                $endgroup$
                – user587054
                Jan 5 at 8:39














              1












              1








              1





              $begingroup$

              Note that for $n^2-n+1$ and $n >1$, the following inequality holds true:



              $$n^2-2n+1 < n^2-n+1 < n^2$$



              $$(n-1)^2 < n^2-n+1 < n^2$$



              where $(n-1)^2$ and $n^2$ are two consecutive squares. What can be concluded from this?






              share|cite|improve this answer











              $endgroup$



              Note that for $n^2-n+1$ and $n >1$, the following inequality holds true:



              $$n^2-2n+1 < n^2-n+1 < n^2$$



              $$(n-1)^2 < n^2-n+1 < n^2$$



              where $(n-1)^2$ and $n^2$ are two consecutive squares. What can be concluded from this?







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 5 at 7:46

























              answered Jan 5 at 7:40









              KM101KM101

              5,8711423




              5,8711423












              • $begingroup$
                Thank you but can you please help me with the second part I updated on the post. Thank you anyways
                $endgroup$
                – user587054
                Jan 5 at 8:39


















              • $begingroup$
                Thank you but can you please help me with the second part I updated on the post. Thank you anyways
                $endgroup$
                – user587054
                Jan 5 at 8:39
















              $begingroup$
              Thank you but can you please help me with the second part I updated on the post. Thank you anyways
              $endgroup$
              – user587054
              Jan 5 at 8:39




              $begingroup$
              Thank you but can you please help me with the second part I updated on the post. Thank you anyways
              $endgroup$
              – user587054
              Jan 5 at 8:39











              0












              $begingroup$

              A hint for the second part. Consider the possibilities for $n pmod3$, that is, $n$ may be of the form $3k$, $3k+1$ or $3k+2$. Do any of these $n$ yield expressions for $2(n^2+1)-n$ that are possible squares $pmod3$? And if any is, is it also a possible square $pmod9$?






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                A hint for the second part. Consider the possibilities for $n pmod3$, that is, $n$ may be of the form $3k$, $3k+1$ or $3k+2$. Do any of these $n$ yield expressions for $2(n^2+1)-n$ that are possible squares $pmod3$? And if any is, is it also a possible square $pmod9$?






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  A hint for the second part. Consider the possibilities for $n pmod3$, that is, $n$ may be of the form $3k$, $3k+1$ or $3k+2$. Do any of these $n$ yield expressions for $2(n^2+1)-n$ that are possible squares $pmod3$? And if any is, is it also a possible square $pmod9$?






                  share|cite|improve this answer









                  $endgroup$



                  A hint for the second part. Consider the possibilities for $n pmod3$, that is, $n$ may be of the form $3k$, $3k+1$ or $3k+2$. Do any of these $n$ yield expressions for $2(n^2+1)-n$ that are possible squares $pmod3$? And if any is, is it also a possible square $pmod9$?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 5 at 12:16









                  Adam BaileyAdam Bailey

                  2,0821319




                  2,0821319






























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