show that no number in the sequence $2, 22, 222, 2222, ldots$ is a perfect square












1












$begingroup$


My question is about the problem show that no number in the sequence 2, 22, 222, 2222, ... is a perfect square.



I am having trouble writing the proof as I am unable to call upon the information to make effective claims.






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    How many times does $2$ divide into such numbers?
    $endgroup$
    – SmileyCraft
    Jan 22 at 19:20






  • 2




    $begingroup$
    Hint: show that all perfect squares are of the form $4n$ or $4n+1$. Note: as an exercise, you can try to prove that no perfect square ends in $2$.
    $endgroup$
    – lulu
    Jan 22 at 19:20








  • 2




    $begingroup$
    @SmileyCraft Why are you answering in a comment?
    $endgroup$
    – Arthur
    Jan 22 at 19:20






  • 2




    $begingroup$
    @lulu Why are you answering in a comment?
    $endgroup$
    – Arthur
    Jan 22 at 19:21






  • 1




    $begingroup$
    @SmileyCraft Yes, it's vague. That's because people like you keep blurring the line. I'm trying to make people aware and conscious about it so maybe the line can be clearer again.
    $endgroup$
    – Arthur
    Jan 22 at 19:25


















1












$begingroup$


My question is about the problem show that no number in the sequence 2, 22, 222, 2222, ... is a perfect square.



I am having trouble writing the proof as I am unable to call upon the information to make effective claims.






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    How many times does $2$ divide into such numbers?
    $endgroup$
    – SmileyCraft
    Jan 22 at 19:20






  • 2




    $begingroup$
    Hint: show that all perfect squares are of the form $4n$ or $4n+1$. Note: as an exercise, you can try to prove that no perfect square ends in $2$.
    $endgroup$
    – lulu
    Jan 22 at 19:20








  • 2




    $begingroup$
    @SmileyCraft Why are you answering in a comment?
    $endgroup$
    – Arthur
    Jan 22 at 19:20






  • 2




    $begingroup$
    @lulu Why are you answering in a comment?
    $endgroup$
    – Arthur
    Jan 22 at 19:21






  • 1




    $begingroup$
    @SmileyCraft Yes, it's vague. That's because people like you keep blurring the line. I'm trying to make people aware and conscious about it so maybe the line can be clearer again.
    $endgroup$
    – Arthur
    Jan 22 at 19:25
















1












1








1





$begingroup$


My question is about the problem show that no number in the sequence 2, 22, 222, 2222, ... is a perfect square.



I am having trouble writing the proof as I am unable to call upon the information to make effective claims.






elementary-number-theory







share|cite|improve this question











$endgroup$




My question is about the problem show that no number in the sequence 2, 22, 222, 2222, ... is a perfect square.



I am having trouble writing the proof as I am unable to call upon the information to make effective claims.






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 22 at 19:20









gt6989b

34.5k22456




34.5k22456










asked Jan 22 at 19:19









Shana RianShana Rian

243




243








  • 3




    $begingroup$
    How many times does $2$ divide into such numbers?
    $endgroup$
    – SmileyCraft
    Jan 22 at 19:20






  • 2




    $begingroup$
    Hint: show that all perfect squares are of the form $4n$ or $4n+1$. Note: as an exercise, you can try to prove that no perfect square ends in $2$.
    $endgroup$
    – lulu
    Jan 22 at 19:20








  • 2




    $begingroup$
    @SmileyCraft Why are you answering in a comment?
    $endgroup$
    – Arthur
    Jan 22 at 19:20






  • 2




    $begingroup$
    @lulu Why are you answering in a comment?
    $endgroup$
    – Arthur
    Jan 22 at 19:21






  • 1




    $begingroup$
    @SmileyCraft Yes, it's vague. That's because people like you keep blurring the line. I'm trying to make people aware and conscious about it so maybe the line can be clearer again.
    $endgroup$
    – Arthur
    Jan 22 at 19:25
















  • 3




    $begingroup$
    How many times does $2$ divide into such numbers?
    $endgroup$
    – SmileyCraft
    Jan 22 at 19:20






  • 2




    $begingroup$
    Hint: show that all perfect squares are of the form $4n$ or $4n+1$. Note: as an exercise, you can try to prove that no perfect square ends in $2$.
    $endgroup$
    – lulu
    Jan 22 at 19:20








  • 2




    $begingroup$
    @SmileyCraft Why are you answering in a comment?
    $endgroup$
    – Arthur
    Jan 22 at 19:20






  • 2




    $begingroup$
    @lulu Why are you answering in a comment?
    $endgroup$
    – Arthur
    Jan 22 at 19:21






  • 1




    $begingroup$
    @SmileyCraft Yes, it's vague. That's because people like you keep blurring the line. I'm trying to make people aware and conscious about it so maybe the line can be clearer again.
    $endgroup$
    – Arthur
    Jan 22 at 19:25










3




3




$begingroup$
How many times does $2$ divide into such numbers?
$endgroup$
– SmileyCraft
Jan 22 at 19:20




$begingroup$
How many times does $2$ divide into such numbers?
$endgroup$
– SmileyCraft
Jan 22 at 19:20




2




2




$begingroup$
Hint: show that all perfect squares are of the form $4n$ or $4n+1$. Note: as an exercise, you can try to prove that no perfect square ends in $2$.
$endgroup$
– lulu
Jan 22 at 19:20






$begingroup$
Hint: show that all perfect squares are of the form $4n$ or $4n+1$. Note: as an exercise, you can try to prove that no perfect square ends in $2$.
$endgroup$
– lulu
Jan 22 at 19:20






2




2




$begingroup$
@SmileyCraft Why are you answering in a comment?
$endgroup$
– Arthur
Jan 22 at 19:20




$begingroup$
@SmileyCraft Why are you answering in a comment?
$endgroup$
– Arthur
Jan 22 at 19:20




2




2




$begingroup$
@lulu Why are you answering in a comment?
$endgroup$
– Arthur
Jan 22 at 19:21




$begingroup$
@lulu Why are you answering in a comment?
$endgroup$
– Arthur
Jan 22 at 19:21




1




1




$begingroup$
@SmileyCraft Yes, it's vague. That's because people like you keep blurring the line. I'm trying to make people aware and conscious about it so maybe the line can be clearer again.
$endgroup$
– Arthur
Jan 22 at 19:25






$begingroup$
@SmileyCraft Yes, it's vague. That's because people like you keep blurring the line. I'm trying to make people aware and conscious about it so maybe the line can be clearer again.
$endgroup$
– Arthur
Jan 22 at 19:25












3 Answers
3






active

oldest

votes


















8












$begingroup$

note any number $N$ in the sequence is divisible by $2$. If $N$ is a square number divisible by $2$, then $N$ is divisible by $4$, and therefore $N/2$ would be even. Clearly for all terms in the sequence $N/2$ ends in $1$ and is therefore odd. We have a contradiction.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Observe that every number in that sequence is congruent to $2 $mod $5$. We can easily compute that, mod $5$, we have
    $$0^{2}equiv 0$$
    $$1^{2}equiv 1$$
    $$2^{2}equiv 4$$
    $$3^{2}equiv 4$$
    $$4^{2}equiv 1$$
    In other words, every square integer is either congruent to $0,1$ or $4$ mod $5$. Therefore it is impossible for a square to be congruent to $2$ mod $5$, which means no square can appear in that sequence.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      If $n$ is a square, then either $nequiv0pmod4$ or $nequiv1pmod4$.



      But of the integers in the sequence, $2equiv2pmod4$, and all the rest are of the form $100m+22$ for some $mgeq0$. In the latter case $100m+22=4(25m+5)+2equiv2pmod4$. So the sequence can contain no square.






      share|cite|improve this answer











      $endgroup$













        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%2f3083584%2fshow-that-no-number-in-the-sequence-2-22-222-2222-ldots-is-a-perfect-squa%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









        8












        $begingroup$

        note any number $N$ in the sequence is divisible by $2$. If $N$ is a square number divisible by $2$, then $N$ is divisible by $4$, and therefore $N/2$ would be even. Clearly for all terms in the sequence $N/2$ ends in $1$ and is therefore odd. We have a contradiction.






        share|cite|improve this answer









        $endgroup$


















          8












          $begingroup$

          note any number $N$ in the sequence is divisible by $2$. If $N$ is a square number divisible by $2$, then $N$ is divisible by $4$, and therefore $N/2$ would be even. Clearly for all terms in the sequence $N/2$ ends in $1$ and is therefore odd. We have a contradiction.






          share|cite|improve this answer









          $endgroup$
















            8












            8








            8





            $begingroup$

            note any number $N$ in the sequence is divisible by $2$. If $N$ is a square number divisible by $2$, then $N$ is divisible by $4$, and therefore $N/2$ would be even. Clearly for all terms in the sequence $N/2$ ends in $1$ and is therefore odd. We have a contradiction.






            share|cite|improve this answer









            $endgroup$



            note any number $N$ in the sequence is divisible by $2$. If $N$ is a square number divisible by $2$, then $N$ is divisible by $4$, and therefore $N/2$ would be even. Clearly for all terms in the sequence $N/2$ ends in $1$ and is therefore odd. We have a contradiction.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 22 at 19:30









            Rohan NuckchadyRohan Nuckchady

            1111




            1111























                2












                $begingroup$

                Observe that every number in that sequence is congruent to $2 $mod $5$. We can easily compute that, mod $5$, we have
                $$0^{2}equiv 0$$
                $$1^{2}equiv 1$$
                $$2^{2}equiv 4$$
                $$3^{2}equiv 4$$
                $$4^{2}equiv 1$$
                In other words, every square integer is either congruent to $0,1$ or $4$ mod $5$. Therefore it is impossible for a square to be congruent to $2$ mod $5$, which means no square can appear in that sequence.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Observe that every number in that sequence is congruent to $2 $mod $5$. We can easily compute that, mod $5$, we have
                  $$0^{2}equiv 0$$
                  $$1^{2}equiv 1$$
                  $$2^{2}equiv 4$$
                  $$3^{2}equiv 4$$
                  $$4^{2}equiv 1$$
                  In other words, every square integer is either congruent to $0,1$ or $4$ mod $5$. Therefore it is impossible for a square to be congruent to $2$ mod $5$, which means no square can appear in that sequence.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Observe that every number in that sequence is congruent to $2 $mod $5$. We can easily compute that, mod $5$, we have
                    $$0^{2}equiv 0$$
                    $$1^{2}equiv 1$$
                    $$2^{2}equiv 4$$
                    $$3^{2}equiv 4$$
                    $$4^{2}equiv 1$$
                    In other words, every square integer is either congruent to $0,1$ or $4$ mod $5$. Therefore it is impossible for a square to be congruent to $2$ mod $5$, which means no square can appear in that sequence.






                    share|cite|improve this answer









                    $endgroup$



                    Observe that every number in that sequence is congruent to $2 $mod $5$. We can easily compute that, mod $5$, we have
                    $$0^{2}equiv 0$$
                    $$1^{2}equiv 1$$
                    $$2^{2}equiv 4$$
                    $$3^{2}equiv 4$$
                    $$4^{2}equiv 1$$
                    In other words, every square integer is either congruent to $0,1$ or $4$ mod $5$. Therefore it is impossible for a square to be congruent to $2$ mod $5$, which means no square can appear in that sequence.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 22 at 19:24









                    pwerthpwerth

                    3,243417




                    3,243417























                        0












                        $begingroup$

                        If $n$ is a square, then either $nequiv0pmod4$ or $nequiv1pmod4$.



                        But of the integers in the sequence, $2equiv2pmod4$, and all the rest are of the form $100m+22$ for some $mgeq0$. In the latter case $100m+22=4(25m+5)+2equiv2pmod4$. So the sequence can contain no square.






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          If $n$ is a square, then either $nequiv0pmod4$ or $nequiv1pmod4$.



                          But of the integers in the sequence, $2equiv2pmod4$, and all the rest are of the form $100m+22$ for some $mgeq0$. In the latter case $100m+22=4(25m+5)+2equiv2pmod4$. So the sequence can contain no square.






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            If $n$ is a square, then either $nequiv0pmod4$ or $nequiv1pmod4$.



                            But of the integers in the sequence, $2equiv2pmod4$, and all the rest are of the form $100m+22$ for some $mgeq0$. In the latter case $100m+22=4(25m+5)+2equiv2pmod4$. So the sequence can contain no square.






                            share|cite|improve this answer











                            $endgroup$



                            If $n$ is a square, then either $nequiv0pmod4$ or $nequiv1pmod4$.



                            But of the integers in the sequence, $2equiv2pmod4$, and all the rest are of the form $100m+22$ for some $mgeq0$. In the latter case $100m+22=4(25m+5)+2equiv2pmod4$. So the sequence can contain no square.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Jan 22 at 19:54

























                            answered Jan 22 at 19:37









                            Adam BaileyAdam Bailey

                            2,1371419




                            2,1371419






























                                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%2f3083584%2fshow-that-no-number-in-the-sequence-2-22-222-2222-ldots-is-a-perfect-squa%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

                                Mario Kart Wii

                                Image
                                Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
                                Read more

                                What does “Dominus providebit” mean?

                                Image
                                5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
                                Read more

                                Antonio Litta Visconti Arese

                                Image
                                Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
                                Read more