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












1












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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







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  • 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






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$begingroup$

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











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        3 Answers
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        3 Answers
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        $begingroup$

        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






























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