Prove: if $c^2+8 equiv 0$ mod $p$ then $c^3-7c^2-8c$ is a quadratic residue mod $p$.Show that if $m$ in $mathbbZ$ has greatest common divisor $1$ with $21$, then $m^6-1$ is divisible by $63$.Prove that if $ab equiv 1 pmodp$ and $a$ is quadratic residue mod $p$, then so is $b$Show that 7 is a quadratic residue for any prime p of the form 28k + 1 and 28k + 3.If p $equiv$ 3 (mod 4) with p prime, prove -1 is a non-quadratic residue modulo p.If $anotequiv 0modp$ then there are $p-1$ solutions (ordered pairs) to $x^2-y^2equiv amodp$Quadratic residue $p equiv 3 pmod4 $Least prime quadratic residueProving that if a is a quadratic residue mod p, then -a is also a quadratic residue mod p iff $p equiv 1 pmod 4$Weird quadratic residue question.Question regarding quadratic residueFinding a non-quadratic residue mod $p$

I’m planning on buying a laser printer but concerned about the life cycle of toner in the machine

How to make payment on the internet without leaving a money trail?

What is the command to reset a PC without deleting any files

How is this relation reflexive?

Email Account under attack (really) - anything I can do?

What typically incentivizes a professor to change jobs to a lower ranking university?

How can I fix this gap between bookcases I made?

Why is an old chain unsafe?

Why linear maps act like matrix multiplication?

The use of multiple foreign keys on same column in SQL Server

A newer friend of my brother's gave him a load of baseball cards that are supposedly extremely valuable. Is this a scam?

GPS Rollover on Android Smartphones

How to report a triplet of septets in NMR tabulation?

How is it possible to have an ability score that is less than 3?

N.B. ligature in Latex

XeLaTeX and pdfLaTeX ignore hyphenation

What do you call a Matrix-like slowdown and camera movement effect?

Why CLRS example on residual networks does not follows its formula?

Motorized valve interfering with button?

Is it possible to make sharp wind that can cut stuff from afar?

How to type dʒ symbol (IPA) on Mac?

Why are only specific transaction types accepted into the mempool?

Copenhagen passport control - US citizen

Is it tax fraud for an individual to declare non-taxable revenue as taxable income? (US tax laws)



Prove: if $c^2+8 equiv 0$ mod $p$ then $c^3-7c^2-8c$ is a quadratic residue mod $p$.


Show that if $m$ in $mathbbZ$ has greatest common divisor $1$ with $21$, then $m^6-1$ is divisible by $63$.Prove that if $ab equiv 1 pmodp$ and $a$ is quadratic residue mod $p$, then so is $b$Show that 7 is a quadratic residue for any prime p of the form 28k + 1 and 28k + 3.If p $equiv$ 3 (mod 4) with p prime, prove -1 is a non-quadratic residue modulo p.If $anotequiv 0modp$ then there are $p-1$ solutions (ordered pairs) to $x^2-y^2equiv amodp$Quadratic residue $p equiv 3 pmod4 $Least prime quadratic residueProving that if a is a quadratic residue mod p, then -a is also a quadratic residue mod p iff $p equiv 1 pmod 4$Weird quadratic residue question.Question regarding quadratic residueFinding a non-quadratic residue mod $p$













6












$begingroup$


I want to show:




If $c^2+8 equiv 0$ mod $p$ for prime $p>3$, then $c^3-7c^2-8c$ is a quadratic residue mod $p$.




I have calculated that $c^3-7c^2-8c equiv -7c^2-16c equiv 56- 16c equiv 8(7-2c) equiv c^2 (2c -7)$, so it should be enough to see that $2c -7$ is a quadratic residue. What now?










share|cite|improve this question











$endgroup$
















    6












    $begingroup$


    I want to show:




    If $c^2+8 equiv 0$ mod $p$ for prime $p>3$, then $c^3-7c^2-8c$ is a quadratic residue mod $p$.




    I have calculated that $c^3-7c^2-8c equiv -7c^2-16c equiv 56- 16c equiv 8(7-2c) equiv c^2 (2c -7)$, so it should be enough to see that $2c -7$ is a quadratic residue. What now?










    share|cite|improve this question











    $endgroup$














      6












      6








      6





      $begingroup$


      I want to show:




      If $c^2+8 equiv 0$ mod $p$ for prime $p>3$, then $c^3-7c^2-8c$ is a quadratic residue mod $p$.




      I have calculated that $c^3-7c^2-8c equiv -7c^2-16c equiv 56- 16c equiv 8(7-2c) equiv c^2 (2c -7)$, so it should be enough to see that $2c -7$ is a quadratic residue. What now?










      share|cite|improve this question











      $endgroup$




      I want to show:




      If $c^2+8 equiv 0$ mod $p$ for prime $p>3$, then $c^3-7c^2-8c$ is a quadratic residue mod $p$.




      I have calculated that $c^3-7c^2-8c equiv -7c^2-16c equiv 56- 16c equiv 8(7-2c) equiv c^2 (2c -7)$, so it should be enough to see that $2c -7$ is a quadratic residue. What now?







      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 Mar 22 at 9:02







      AJ.

















      asked Mar 22 at 8:46









      AJ.AJ.

      333




      333




















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          $$2c-7equiv2c-7+c^2+8pmod pequiv(c+1)^2$$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            @AJ. Have you noticed my other answer. That is much more natural derivation
            $endgroup$
            – lab bhattacharjee
            Mar 22 at 9:10


















          3












          $begingroup$

          We have that
          $$(c +1)^2 = c^2 + 2c + 1 = c^2 + 2c + 8 -7 = (c^2 + 8) + (2c - 7) equiv 2c - 7 mod p.$$
          This proves the claim.






          share|cite|improve this answer









          $endgroup$




















            2












            $begingroup$

            Hint:



            As $c^2equiv-8pmod p,$



            $$c(c+1)=c^2+cequiv c-8pmod p$$



            $c(c+1)(c-8)equiv?$



            I believe this is how the problem naturally came into being .






            share|cite|improve this answer









            $endgroup$








            • 1




              $begingroup$
              Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different.
              $endgroup$
              – lab bhattacharjee
              Mar 22 at 9:21










            • $begingroup$
              Me neither... $(+1)$
              $endgroup$
              – user477343
              Mar 24 at 6:12











            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%2f3157905%2fprove-if-c28-equiv-0-mod-p-then-c3-7c2-8c-is-a-quadratic-residue-mod%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            $$2c-7equiv2c-7+c^2+8pmod pequiv(c+1)^2$$






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              @AJ. Have you noticed my other answer. That is much more natural derivation
              $endgroup$
              – lab bhattacharjee
              Mar 22 at 9:10















            4












            $begingroup$

            $$2c-7equiv2c-7+c^2+8pmod pequiv(c+1)^2$$






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              @AJ. Have you noticed my other answer. That is much more natural derivation
              $endgroup$
              – lab bhattacharjee
              Mar 22 at 9:10













            4












            4








            4





            $begingroup$

            $$2c-7equiv2c-7+c^2+8pmod pequiv(c+1)^2$$






            share|cite|improve this answer









            $endgroup$



            $$2c-7equiv2c-7+c^2+8pmod pequiv(c+1)^2$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 22 at 8:53









            lab bhattacharjeelab bhattacharjee

            228k15158279




            228k15158279











            • $begingroup$
              @AJ. Have you noticed my other answer. That is much more natural derivation
              $endgroup$
              – lab bhattacharjee
              Mar 22 at 9:10
















            • $begingroup$
              @AJ. Have you noticed my other answer. That is much more natural derivation
              $endgroup$
              – lab bhattacharjee
              Mar 22 at 9:10















            $begingroup$
            @AJ. Have you noticed my other answer. That is much more natural derivation
            $endgroup$
            – lab bhattacharjee
            Mar 22 at 9:10




            $begingroup$
            @AJ. Have you noticed my other answer. That is much more natural derivation
            $endgroup$
            – lab bhattacharjee
            Mar 22 at 9:10











            3












            $begingroup$

            We have that
            $$(c +1)^2 = c^2 + 2c + 1 = c^2 + 2c + 8 -7 = (c^2 + 8) + (2c - 7) equiv 2c - 7 mod p.$$
            This proves the claim.






            share|cite|improve this answer









            $endgroup$

















              3












              $begingroup$

              We have that
              $$(c +1)^2 = c^2 + 2c + 1 = c^2 + 2c + 8 -7 = (c^2 + 8) + (2c - 7) equiv 2c - 7 mod p.$$
              This proves the claim.






              share|cite|improve this answer









              $endgroup$















                3












                3








                3





                $begingroup$

                We have that
                $$(c +1)^2 = c^2 + 2c + 1 = c^2 + 2c + 8 -7 = (c^2 + 8) + (2c - 7) equiv 2c - 7 mod p.$$
                This proves the claim.






                share|cite|improve this answer









                $endgroup$



                We have that
                $$(c +1)^2 = c^2 + 2c + 1 = c^2 + 2c + 8 -7 = (c^2 + 8) + (2c - 7) equiv 2c - 7 mod p.$$
                This proves the claim.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 22 at 8:54









                StudentStudent

                2,1881727




                2,1881727





















                    2












                    $begingroup$

                    Hint:



                    As $c^2equiv-8pmod p,$



                    $$c(c+1)=c^2+cequiv c-8pmod p$$



                    $c(c+1)(c-8)equiv?$



                    I believe this is how the problem naturally came into being .






                    share|cite|improve this answer









                    $endgroup$








                    • 1




                      $begingroup$
                      Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different.
                      $endgroup$
                      – lab bhattacharjee
                      Mar 22 at 9:21










                    • $begingroup$
                      Me neither... $(+1)$
                      $endgroup$
                      – user477343
                      Mar 24 at 6:12















                    2












                    $begingroup$

                    Hint:



                    As $c^2equiv-8pmod p,$



                    $$c(c+1)=c^2+cequiv c-8pmod p$$



                    $c(c+1)(c-8)equiv?$



                    I believe this is how the problem naturally came into being .






                    share|cite|improve this answer









                    $endgroup$








                    • 1




                      $begingroup$
                      Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different.
                      $endgroup$
                      – lab bhattacharjee
                      Mar 22 at 9:21










                    • $begingroup$
                      Me neither... $(+1)$
                      $endgroup$
                      – user477343
                      Mar 24 at 6:12













                    2












                    2








                    2





                    $begingroup$

                    Hint:



                    As $c^2equiv-8pmod p,$



                    $$c(c+1)=c^2+cequiv c-8pmod p$$



                    $c(c+1)(c-8)equiv?$



                    I believe this is how the problem naturally came into being .






                    share|cite|improve this answer









                    $endgroup$



                    Hint:



                    As $c^2equiv-8pmod p,$



                    $$c(c+1)=c^2+cequiv c-8pmod p$$



                    $c(c+1)(c-8)equiv?$



                    I believe this is how the problem naturally came into being .







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 22 at 8:57









                    lab bhattacharjeelab bhattacharjee

                    228k15158279




                    228k15158279







                    • 1




                      $begingroup$
                      Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different.
                      $endgroup$
                      – lab bhattacharjee
                      Mar 22 at 9:21










                    • $begingroup$
                      Me neither... $(+1)$
                      $endgroup$
                      – user477343
                      Mar 24 at 6:12












                    • 1




                      $begingroup$
                      Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different.
                      $endgroup$
                      – lab bhattacharjee
                      Mar 22 at 9:21










                    • $begingroup$
                      Me neither... $(+1)$
                      $endgroup$
                      – user477343
                      Mar 24 at 6:12







                    1




                    1




                    $begingroup$
                    Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different.
                    $endgroup$
                    – lab bhattacharjee
                    Mar 22 at 9:21




                    $begingroup$
                    Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different.
                    $endgroup$
                    – lab bhattacharjee
                    Mar 22 at 9:21












                    $begingroup$
                    Me neither... $(+1)$
                    $endgroup$
                    – user477343
                    Mar 24 at 6:12




                    $begingroup$
                    Me neither... $(+1)$
                    $endgroup$
                    – user477343
                    Mar 24 at 6:12

















                    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%2f3157905%2fprove-if-c28-equiv-0-mod-p-then-c3-7c2-8c-is-a-quadratic-residue-mod%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

                    How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer

                    random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

                    Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye