proof that $frac{m^a}{2}+frac{m}{2}-1$ is not a prime number












3












$begingroup$


I'm having trouble proving the following:




If $m$ is an even number, $mge4$ and $a$ is an integer, $age2$ then $$frac{m^a}{2}+frac{m}{2}-1$$ is not a prime number.




Usually, in this type of exercises, someone has to show that the number can be written as a product of more than two expressions but whatever I've tried doesn't work...Thank you!






number-theory elementary-number-theory prime-numbers







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I'm having trouble proving the following:




    If $m$ is an even number, $mge4$ and $a$ is an integer, $age2$ then $$frac{m^a}{2}+frac{m}{2}-1$$ is not a prime number.




    Usually, in this type of exercises, someone has to show that the number can be written as a product of more than two expressions but whatever I've tried doesn't work...Thank you!






    number-theory elementary-number-theory prime-numbers







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      2



      $begingroup$


      I'm having trouble proving the following:




      If $m$ is an even number, $mge4$ and $a$ is an integer, $age2$ then $$frac{m^a}{2}+frac{m}{2}-1$$ is not a prime number.




      Usually, in this type of exercises, someone has to show that the number can be written as a product of more than two expressions but whatever I've tried doesn't work...Thank you!






      number-theory elementary-number-theory prime-numbers







      share|cite|improve this question











      $endgroup$




      I'm having trouble proving the following:




      If $m$ is an even number, $mge4$ and $a$ is an integer, $age2$ then $$frac{m^a}{2}+frac{m}{2}-1$$ is not a prime number.




      Usually, in this type of exercises, someone has to show that the number can be written as a product of more than two expressions but whatever I've tried doesn't work...Thank you!






      number-theory elementary-number-theory prime-numbers




      number-theory elementary-number-theory prime-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 15:18









      greedoid

      43.2k1153105




      43.2k1153105










      asked Dec 13 '18 at 13:43









      hlx55555hlx55555

      313




      313






















          2 Answers
          2






          active

          oldest

          votes


















          6












          $begingroup$

          $${m^aover 2}+{mover 2}-1 = {m^a+m-2over 2} = {m^a-1 + m-1over 2} $$



          $$= {(m-1)(m^{a-1} + ...+m+1)+(m-1)over 2} $$



          $$= underbrace{(m-1)}_{=a}underbrace{m^{a-1} + ...+m+2over 2}_{=b} $$



          Clearly $a>1$ and $b>1$ and $b$ is an integer, since $m$ is even.



          So, the number is product of two numbers both $>1$, so it is not a prime.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            so my number is divided by 2, therefore my number is not a prime... right? thank you!
            $endgroup$
            – hlx55555
            Dec 13 '18 at 14:01












          • $begingroup$
            No, your number is product of two numbers both >1, so it is not a prime.
            $endgroup$
            – greedoid
            Dec 13 '18 at 14:02






          • 1




            $begingroup$
            any number divided by 2 (except 2 itself) is not a prime number
            $endgroup$
            – Reinstein
            Dec 13 '18 at 14:02






          • 1




            $begingroup$
            @hlx55555 Try $m=4$ and $a=2$. The number doesn't have to be even.
            $endgroup$
            – yurnero
            Dec 13 '18 at 14:06






          • 1




            $begingroup$
            @yurnero ok, now I see what you both mean... the mistake I made was looking at the numerator and not the number as a whole. When the numerator is divided by two, the quotient is not necessarily an even number
            $endgroup$
            – hlx55555
            Dec 13 '18 at 14:15





















          4












          $begingroup$

          Hint:



          $$m^a+m-2$$ is divisible by $m-1$ and is even as $m^a+m=m(1+m^{a-1})$ is even for integer $m$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            wow, such quick response!! Thank you so much! I can't believe it was such a simple answer...
            $endgroup$
            – hlx55555
            Dec 13 '18 at 13:52










          • $begingroup$
            @hlx55555 "I begin to think, Watson," said Holmes, "that I make a mistake in explaining. 'Omne ignotum pro magnifico,' you know, and my poor little reputation, such as it is, will suffer shipwreck if I am so candid" --- Arthur Conan Doyle in Red Headed League (1891).
            $endgroup$
            – yurnero
            Dec 13 '18 at 13:57











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038017%2fproof-that-fracma2-fracm2-1-is-not-a-prime-number%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          $${m^aover 2}+{mover 2}-1 = {m^a+m-2over 2} = {m^a-1 + m-1over 2} $$



          $$= {(m-1)(m^{a-1} + ...+m+1)+(m-1)over 2} $$



          $$= underbrace{(m-1)}_{=a}underbrace{m^{a-1} + ...+m+2over 2}_{=b} $$



          Clearly $a>1$ and $b>1$ and $b$ is an integer, since $m$ is even.



          So, the number is product of two numbers both $>1$, so it is not a prime.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            so my number is divided by 2, therefore my number is not a prime... right? thank you!
            $endgroup$
            – hlx55555
            Dec 13 '18 at 14:01












          • $begingroup$
            No, your number is product of two numbers both >1, so it is not a prime.
            $endgroup$
            – greedoid
            Dec 13 '18 at 14:02






          • 1




            $begingroup$
            any number divided by 2 (except 2 itself) is not a prime number
            $endgroup$
            – Reinstein
            Dec 13 '18 at 14:02






          • 1




            $begingroup$
            @hlx55555 Try $m=4$ and $a=2$. The number doesn't have to be even.
            $endgroup$
            – yurnero
            Dec 13 '18 at 14:06






          • 1




            $begingroup$
            @yurnero ok, now I see what you both mean... the mistake I made was looking at the numerator and not the number as a whole. When the numerator is divided by two, the quotient is not necessarily an even number
            $endgroup$
            – hlx55555
            Dec 13 '18 at 14:15


















          6












          $begingroup$

          $${m^aover 2}+{mover 2}-1 = {m^a+m-2over 2} = {m^a-1 + m-1over 2} $$



          $$= {(m-1)(m^{a-1} + ...+m+1)+(m-1)over 2} $$



          $$= underbrace{(m-1)}_{=a}underbrace{m^{a-1} + ...+m+2over 2}_{=b} $$



          Clearly $a>1$ and $b>1$ and $b$ is an integer, since $m$ is even.



          So, the number is product of two numbers both $>1$, so it is not a prime.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            so my number is divided by 2, therefore my number is not a prime... right? thank you!
            $endgroup$
            – hlx55555
            Dec 13 '18 at 14:01












          • $begingroup$
            No, your number is product of two numbers both >1, so it is not a prime.
            $endgroup$
            – greedoid
            Dec 13 '18 at 14:02






          • 1




            $begingroup$
            any number divided by 2 (except 2 itself) is not a prime number
            $endgroup$
            – Reinstein
            Dec 13 '18 at 14:02






          • 1




            $begingroup$
            @hlx55555 Try $m=4$ and $a=2$. The number doesn't have to be even.
            $endgroup$
            – yurnero
            Dec 13 '18 at 14:06






          • 1




            $begingroup$
            @yurnero ok, now I see what you both mean... the mistake I made was looking at the numerator and not the number as a whole. When the numerator is divided by two, the quotient is not necessarily an even number
            $endgroup$
            – hlx55555
            Dec 13 '18 at 14:15
















          6












          6








          6





          $begingroup$

          $${m^aover 2}+{mover 2}-1 = {m^a+m-2over 2} = {m^a-1 + m-1over 2} $$



          $$= {(m-1)(m^{a-1} + ...+m+1)+(m-1)over 2} $$



          $$= underbrace{(m-1)}_{=a}underbrace{m^{a-1} + ...+m+2over 2}_{=b} $$



          Clearly $a>1$ and $b>1$ and $b$ is an integer, since $m$ is even.



          So, the number is product of two numbers both $>1$, so it is not a prime.






          share|cite|improve this answer











          $endgroup$



          $${m^aover 2}+{mover 2}-1 = {m^a+m-2over 2} = {m^a-1 + m-1over 2} $$



          $$= {(m-1)(m^{a-1} + ...+m+1)+(m-1)over 2} $$



          $$= underbrace{(m-1)}_{=a}underbrace{m^{a-1} + ...+m+2over 2}_{=b} $$



          Clearly $a>1$ and $b>1$ and $b$ is an integer, since $m$ is even.



          So, the number is product of two numbers both $>1$, so it is not a prime.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 13 '18 at 14:04

























          answered Dec 13 '18 at 13:47









          greedoidgreedoid

          43.2k1153105




          43.2k1153105












          • $begingroup$
            so my number is divided by 2, therefore my number is not a prime... right? thank you!
            $endgroup$
            – hlx55555
            Dec 13 '18 at 14:01












          • $begingroup$
            No, your number is product of two numbers both >1, so it is not a prime.
            $endgroup$
            – greedoid
            Dec 13 '18 at 14:02






          • 1




            $begingroup$
            any number divided by 2 (except 2 itself) is not a prime number
            $endgroup$
            – Reinstein
            Dec 13 '18 at 14:02






          • 1




            $begingroup$
            @hlx55555 Try $m=4$ and $a=2$. The number doesn't have to be even.
            $endgroup$
            – yurnero
            Dec 13 '18 at 14:06






          • 1




            $begingroup$
            @yurnero ok, now I see what you both mean... the mistake I made was looking at the numerator and not the number as a whole. When the numerator is divided by two, the quotient is not necessarily an even number
            $endgroup$
            – hlx55555
            Dec 13 '18 at 14:15




















          • $begingroup$
            so my number is divided by 2, therefore my number is not a prime... right? thank you!
            $endgroup$
            – hlx55555
            Dec 13 '18 at 14:01












          • $begingroup$
            No, your number is product of two numbers both >1, so it is not a prime.
            $endgroup$
            – greedoid
            Dec 13 '18 at 14:02






          • 1




            $begingroup$
            any number divided by 2 (except 2 itself) is not a prime number
            $endgroup$
            – Reinstein
            Dec 13 '18 at 14:02






          • 1




            $begingroup$
            @hlx55555 Try $m=4$ and $a=2$. The number doesn't have to be even.
            $endgroup$
            – yurnero
            Dec 13 '18 at 14:06






          • 1




            $begingroup$
            @yurnero ok, now I see what you both mean... the mistake I made was looking at the numerator and not the number as a whole. When the numerator is divided by two, the quotient is not necessarily an even number
            $endgroup$
            – hlx55555
            Dec 13 '18 at 14:15


















          $begingroup$
          so my number is divided by 2, therefore my number is not a prime... right? thank you!
          $endgroup$
          – hlx55555
          Dec 13 '18 at 14:01






          $begingroup$
          so my number is divided by 2, therefore my number is not a prime... right? thank you!
          $endgroup$
          – hlx55555
          Dec 13 '18 at 14:01














          $begingroup$
          No, your number is product of two numbers both >1, so it is not a prime.
          $endgroup$
          – greedoid
          Dec 13 '18 at 14:02




          $begingroup$
          No, your number is product of two numbers both >1, so it is not a prime.
          $endgroup$
          – greedoid
          Dec 13 '18 at 14:02




          1




          1




          $begingroup$
          any number divided by 2 (except 2 itself) is not a prime number
          $endgroup$
          – Reinstein
          Dec 13 '18 at 14:02




          $begingroup$
          any number divided by 2 (except 2 itself) is not a prime number
          $endgroup$
          – Reinstein
          Dec 13 '18 at 14:02




          1




          1




          $begingroup$
          @hlx55555 Try $m=4$ and $a=2$. The number doesn't have to be even.
          $endgroup$
          – yurnero
          Dec 13 '18 at 14:06




          $begingroup$
          @hlx55555 Try $m=4$ and $a=2$. The number doesn't have to be even.
          $endgroup$
          – yurnero
          Dec 13 '18 at 14:06




          1




          1




          $begingroup$
          @yurnero ok, now I see what you both mean... the mistake I made was looking at the numerator and not the number as a whole. When the numerator is divided by two, the quotient is not necessarily an even number
          $endgroup$
          – hlx55555
          Dec 13 '18 at 14:15






          $begingroup$
          @yurnero ok, now I see what you both mean... the mistake I made was looking at the numerator and not the number as a whole. When the numerator is divided by two, the quotient is not necessarily an even number
          $endgroup$
          – hlx55555
          Dec 13 '18 at 14:15













          4












          $begingroup$

          Hint:



          $$m^a+m-2$$ is divisible by $m-1$ and is even as $m^a+m=m(1+m^{a-1})$ is even for integer $m$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            wow, such quick response!! Thank you so much! I can't believe it was such a simple answer...
            $endgroup$
            – hlx55555
            Dec 13 '18 at 13:52










          • $begingroup$
            @hlx55555 "I begin to think, Watson," said Holmes, "that I make a mistake in explaining. 'Omne ignotum pro magnifico,' you know, and my poor little reputation, such as it is, will suffer shipwreck if I am so candid" --- Arthur Conan Doyle in Red Headed League (1891).
            $endgroup$
            – yurnero
            Dec 13 '18 at 13:57
















          4












          $begingroup$

          Hint:



          $$m^a+m-2$$ is divisible by $m-1$ and is even as $m^a+m=m(1+m^{a-1})$ is even for integer $m$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            wow, such quick response!! Thank you so much! I can't believe it was such a simple answer...
            $endgroup$
            – hlx55555
            Dec 13 '18 at 13:52










          • $begingroup$
            @hlx55555 "I begin to think, Watson," said Holmes, "that I make a mistake in explaining. 'Omne ignotum pro magnifico,' you know, and my poor little reputation, such as it is, will suffer shipwreck if I am so candid" --- Arthur Conan Doyle in Red Headed League (1891).
            $endgroup$
            – yurnero
            Dec 13 '18 at 13:57














          4












          4








          4





          $begingroup$

          Hint:



          $$m^a+m-2$$ is divisible by $m-1$ and is even as $m^a+m=m(1+m^{a-1})$ is even for integer $m$






          share|cite|improve this answer









          $endgroup$



          Hint:



          $$m^a+m-2$$ is divisible by $m-1$ and is even as $m^a+m=m(1+m^{a-1})$ is even for integer $m$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 13:48









          lab bhattacharjeelab bhattacharjee

          226k15157275




          226k15157275












          • $begingroup$
            wow, such quick response!! Thank you so much! I can't believe it was such a simple answer...
            $endgroup$
            – hlx55555
            Dec 13 '18 at 13:52










          • $begingroup$
            @hlx55555 "I begin to think, Watson," said Holmes, "that I make a mistake in explaining. 'Omne ignotum pro magnifico,' you know, and my poor little reputation, such as it is, will suffer shipwreck if I am so candid" --- Arthur Conan Doyle in Red Headed League (1891).
            $endgroup$
            – yurnero
            Dec 13 '18 at 13:57


















          • $begingroup$
            wow, such quick response!! Thank you so much! I can't believe it was such a simple answer...
            $endgroup$
            – hlx55555
            Dec 13 '18 at 13:52










          • $begingroup$
            @hlx55555 "I begin to think, Watson," said Holmes, "that I make a mistake in explaining. 'Omne ignotum pro magnifico,' you know, and my poor little reputation, such as it is, will suffer shipwreck if I am so candid" --- Arthur Conan Doyle in Red Headed League (1891).
            $endgroup$
            – yurnero
            Dec 13 '18 at 13:57
















          $begingroup$
          wow, such quick response!! Thank you so much! I can't believe it was such a simple answer...
          $endgroup$
          – hlx55555
          Dec 13 '18 at 13:52




          $begingroup$
          wow, such quick response!! Thank you so much! I can't believe it was such a simple answer...
          $endgroup$
          – hlx55555
          Dec 13 '18 at 13:52












          $begingroup$
          @hlx55555 "I begin to think, Watson," said Holmes, "that I make a mistake in explaining. 'Omne ignotum pro magnifico,' you know, and my poor little reputation, such as it is, will suffer shipwreck if I am so candid" --- Arthur Conan Doyle in Red Headed League (1891).
          $endgroup$
          – yurnero
          Dec 13 '18 at 13:57




          $begingroup$
          @hlx55555 "I begin to think, Watson," said Holmes, "that I make a mistake in explaining. 'Omne ignotum pro magnifico,' you know, and my poor little reputation, such as it is, will suffer shipwreck if I am so candid" --- Arthur Conan Doyle in Red Headed League (1891).
          $endgroup$
          – yurnero
          Dec 13 '18 at 13:57


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038017%2fproof-that-fracma2-fracm2-1-is-not-a-prime-number%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Probability when a professor distributes a quiz and homework assignment to a class of n students.

          Image
          0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
          Read more

          Aardman Animations

          Image
          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
          Read more

          Are they similar matrix

          Image
          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
          Read more