Prove that for all integers $n$ if $3 mid n^2$, then $3 mid n$












0












$begingroup$


Prove that for all entegers $n$ if $3$ | $n^2$, then $3$ | $n$.



I figured using contropositve was the best method by using the definition "an integer $k$ is not divisible by 3 if and only if there exists an integer $k$ such that $n=3k+1$ or $n=3k+2$. Also using the definition $a$ divides $b$ written $a$|$b$ if $b=ac$ for some $c$ in integers.



Here it goes:



$$3nmid n$$



$$3k+1=3n$$



$$frac{3k+1}{3}=n$$






proof-verification proof-writing proof-explanation







share|cite|improve this question











$endgroup$












  • $begingroup$
    Use Euclid's lemma, since $3$ is prime.
    $endgroup$
    – rtybase
    Jan 20 at 2:01










  • $begingroup$
    I assume Euclid's lemma hasn't be introduced yet. As that would make this trivial.
    $endgroup$
    – fleablood
    Jan 20 at 2:05






  • 1




    $begingroup$
    Why does $3not mid n$ mean $3k + 1 = 3n$??????
    $endgroup$
    – fleablood
    Jan 20 at 2:06










  • $begingroup$
    Yes we have not gotten to Euclid's lemma. In a month we are covering that. For now the question is in the chapter of contrapositive and contradiction.
    $endgroup$
    – John
    Jan 20 at 2:09






  • 2




    $begingroup$
    The reason I am asking is, you mentioned $n=3k+1$ or $n=3k+2$ so you are familiar with the divisibility with remainder theorem?
    $endgroup$
    – rtybase
    Jan 20 at 2:14


















0












$begingroup$


Prove that for all entegers $n$ if $3$ | $n^2$, then $3$ | $n$.



I figured using contropositve was the best method by using the definition "an integer $k$ is not divisible by 3 if and only if there exists an integer $k$ such that $n=3k+1$ or $n=3k+2$. Also using the definition $a$ divides $b$ written $a$|$b$ if $b=ac$ for some $c$ in integers.



Here it goes:



$$3nmid n$$



$$3k+1=3n$$



$$frac{3k+1}{3}=n$$






proof-verification proof-writing proof-explanation







share|cite|improve this question











$endgroup$












  • $begingroup$
    Use Euclid's lemma, since $3$ is prime.
    $endgroup$
    – rtybase
    Jan 20 at 2:01










  • $begingroup$
    I assume Euclid's lemma hasn't be introduced yet. As that would make this trivial.
    $endgroup$
    – fleablood
    Jan 20 at 2:05






  • 1




    $begingroup$
    Why does $3not mid n$ mean $3k + 1 = 3n$??????
    $endgroup$
    – fleablood
    Jan 20 at 2:06










  • $begingroup$
    Yes we have not gotten to Euclid's lemma. In a month we are covering that. For now the question is in the chapter of contrapositive and contradiction.
    $endgroup$
    – John
    Jan 20 at 2:09






  • 2




    $begingroup$
    The reason I am asking is, you mentioned $n=3k+1$ or $n=3k+2$ so you are familiar with the divisibility with remainder theorem?
    $endgroup$
    – rtybase
    Jan 20 at 2:14
















0












0








0


1



$begingroup$


Prove that for all entegers $n$ if $3$ | $n^2$, then $3$ | $n$.



I figured using contropositve was the best method by using the definition "an integer $k$ is not divisible by 3 if and only if there exists an integer $k$ such that $n=3k+1$ or $n=3k+2$. Also using the definition $a$ divides $b$ written $a$|$b$ if $b=ac$ for some $c$ in integers.



Here it goes:



$$3nmid n$$



$$3k+1=3n$$



$$frac{3k+1}{3}=n$$






proof-verification proof-writing proof-explanation







share|cite|improve this question











$endgroup$




Prove that for all entegers $n$ if $3$ | $n^2$, then $3$ | $n$.



I figured using contropositve was the best method by using the definition "an integer $k$ is not divisible by 3 if and only if there exists an integer $k$ such that $n=3k+1$ or $n=3k+2$. Also using the definition $a$ divides $b$ written $a$|$b$ if $b=ac$ for some $c$ in integers.



Here it goes:



$$3nmid n$$



$$3k+1=3n$$



$$frac{3k+1}{3}=n$$






proof-verification proof-writing proof-explanation




proof-verification proof-writing proof-explanation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 20 at 2:04









rtybase

11k21533




11k21533










asked Jan 20 at 1:52









JohnJohn

325




325












  • $begingroup$
    Use Euclid's lemma, since $3$ is prime.
    $endgroup$
    – rtybase
    Jan 20 at 2:01










  • $begingroup$
    I assume Euclid's lemma hasn't be introduced yet. As that would make this trivial.
    $endgroup$
    – fleablood
    Jan 20 at 2:05






  • 1




    $begingroup$
    Why does $3not mid n$ mean $3k + 1 = 3n$??????
    $endgroup$
    – fleablood
    Jan 20 at 2:06










  • $begingroup$
    Yes we have not gotten to Euclid's lemma. In a month we are covering that. For now the question is in the chapter of contrapositive and contradiction.
    $endgroup$
    – John
    Jan 20 at 2:09






  • 2




    $begingroup$
    The reason I am asking is, you mentioned $n=3k+1$ or $n=3k+2$ so you are familiar with the divisibility with remainder theorem?
    $endgroup$
    – rtybase
    Jan 20 at 2:14




















  • $begingroup$
    Use Euclid's lemma, since $3$ is prime.
    $endgroup$
    – rtybase
    Jan 20 at 2:01










  • $begingroup$
    I assume Euclid's lemma hasn't be introduced yet. As that would make this trivial.
    $endgroup$
    – fleablood
    Jan 20 at 2:05






  • 1




    $begingroup$
    Why does $3not mid n$ mean $3k + 1 = 3n$??????
    $endgroup$
    – fleablood
    Jan 20 at 2:06










  • $begingroup$
    Yes we have not gotten to Euclid's lemma. In a month we are covering that. For now the question is in the chapter of contrapositive and contradiction.
    $endgroup$
    – John
    Jan 20 at 2:09






  • 2




    $begingroup$
    The reason I am asking is, you mentioned $n=3k+1$ or $n=3k+2$ so you are familiar with the divisibility with remainder theorem?
    $endgroup$
    – rtybase
    Jan 20 at 2:14


















$begingroup$
Use Euclid's lemma, since $3$ is prime.
$endgroup$
– rtybase
Jan 20 at 2:01




$begingroup$
Use Euclid's lemma, since $3$ is prime.
$endgroup$
– rtybase
Jan 20 at 2:01












$begingroup$
I assume Euclid's lemma hasn't be introduced yet. As that would make this trivial.
$endgroup$
– fleablood
Jan 20 at 2:05




$begingroup$
I assume Euclid's lemma hasn't be introduced yet. As that would make this trivial.
$endgroup$
– fleablood
Jan 20 at 2:05




1




1




$begingroup$
Why does $3not mid n$ mean $3k + 1 = 3n$??????
$endgroup$
– fleablood
Jan 20 at 2:06




$begingroup$
Why does $3not mid n$ mean $3k + 1 = 3n$??????
$endgroup$
– fleablood
Jan 20 at 2:06












$begingroup$
Yes we have not gotten to Euclid's lemma. In a month we are covering that. For now the question is in the chapter of contrapositive and contradiction.
$endgroup$
– John
Jan 20 at 2:09




$begingroup$
Yes we have not gotten to Euclid's lemma. In a month we are covering that. For now the question is in the chapter of contrapositive and contradiction.
$endgroup$
– John
Jan 20 at 2:09




2




2




$begingroup$
The reason I am asking is, you mentioned $n=3k+1$ or $n=3k+2$ so you are familiar with the divisibility with remainder theorem?
$endgroup$
– rtybase
Jan 20 at 2:14






$begingroup$
The reason I am asking is, you mentioned $n=3k+1$ or $n=3k+2$ so you are familiar with the divisibility with remainder theorem?
$endgroup$
– rtybase
Jan 20 at 2:14












4 Answers
4






active

oldest

votes


















0












$begingroup$

I believe a simpler way is to use the unique factorization to show that for



$$n = prod_{i , = , 1}^{m} p_i^{a_i} tag{1}label{eq1}$$



where the $p_i$ are unique primes, means that



$$n^2 = prod_{i , = , 1}^{m} p_i^{2a_i} tag{2}label{eq2}$$



Thus, if $3 ; vert ; n^2$, then one of the $p_i$ must be $3$, so $3 ; vert ; n$.



Also, using your suggestion, if $n = 3k + 1$ or $n = 3k + 2$, then $n^2 = 9k^2 + 6k + 1$ or $n^2 = 9k^2 + 12k + 4$. In either case, when dividing by $3$, there is a remainder of $1$, showing that $3$ doesn't divide $n^2$. But, as $3 ; vert ; n^2$, then $n$ cannot be either of the $2$ forms, so it must be $n = 3k$, giving that $3 ; vert ; n$. This is an example of proving what's requested using contrapositive as it shows that if $3 not{vert} ; n$, then $3 not{vert} ; n^2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for you comment. We have not gotten to unique factorization yet in mathematical reasoning class. Would my proof suffice?
    $endgroup$
    – John
    Jan 20 at 2:00










  • $begingroup$
    @John I was editing my response when you wrote your comment. I hope it sufficiently answers your question.
    $endgroup$
    – John Omielan
    Jan 20 at 2:02










  • $begingroup$
    @John Based on the definition for contrapositive, I believe what I showed in the last paragraph uses the contrapositive technique. As for what you have in your question, as one of the comments state, I'm not sure how you get that $3k + 1 = 3n$, i.e., where does the "$3$" part come from in $3n$?
    $endgroup$
    – John Omielan
    Jan 20 at 2:17



















0












$begingroup$

Presumably you have had the division theorem.



For $n$ and integer there exist integer $k, r$ so that $n = 3k + r$ where $0 le r < 3$. So $r = 0, 1$ or $2$.



Can you accept that?



If $r=0$ then $3|n$.



If $r = 1$ then $n^2 = (3k + 1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1$ and $3not mid n^2$ and that's a contradiction.



If $r = 2$ then $n^2= (3k + 2)^2 = 9k^2 + 12k + 4 = 3(k^2 + 4k+ 1) + 1$ and $3not mid n^2$ and that's a contradiction.



So if $3|n^2$ then the only possibility is $n = 3k$ for some $k$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You don't need to resort to proving the contrapositive; it's possible to prove the statement directly:



    If we take as given that $3$ divides (exactly) one of the three consecutive numbers $n-1$, $n$, and $n+1$, then $3$ divides their product, $(n-1)n(n+1)=n^3-n$. Now if $3$ divides $n^2$, then it also divides $n^3$, and thus it divides the difference, $n^3-(n^3-n)=n$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      In
      If $n mid a^2 $, what is the largest $m$ for which $m mid a$?,
      I prove this result:



      Given
      $n = prod p_i^{a_i}$,
      then the largest $m$
      such that
      $m | a$ for all $a$
      such that
      $n | a^2$ is
      $m = prod p_i^{lceil frac{a_i}{2}rceil}$.



      If $n$ is a prime,
      $3$ in this problems case,
      then
      $n = 3^1$
      so
      $m = 3^1$
      also.






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        0












        $begingroup$

        I believe a simpler way is to use the unique factorization to show that for



        $$n = prod_{i , = , 1}^{m} p_i^{a_i} tag{1}label{eq1}$$



        where the $p_i$ are unique primes, means that



        $$n^2 = prod_{i , = , 1}^{m} p_i^{2a_i} tag{2}label{eq2}$$



        Thus, if $3 ; vert ; n^2$, then one of the $p_i$ must be $3$, so $3 ; vert ; n$.



        Also, using your suggestion, if $n = 3k + 1$ or $n = 3k + 2$, then $n^2 = 9k^2 + 6k + 1$ or $n^2 = 9k^2 + 12k + 4$. In either case, when dividing by $3$, there is a remainder of $1$, showing that $3$ doesn't divide $n^2$. But, as $3 ; vert ; n^2$, then $n$ cannot be either of the $2$ forms, so it must be $n = 3k$, giving that $3 ; vert ; n$. This is an example of proving what's requested using contrapositive as it shows that if $3 not{vert} ; n$, then $3 not{vert} ; n^2$.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Thanks for you comment. We have not gotten to unique factorization yet in mathematical reasoning class. Would my proof suffice?
          $endgroup$
          – John
          Jan 20 at 2:00










        • $begingroup$
          @John I was editing my response when you wrote your comment. I hope it sufficiently answers your question.
          $endgroup$
          – John Omielan
          Jan 20 at 2:02










        • $begingroup$
          @John Based on the definition for contrapositive, I believe what I showed in the last paragraph uses the contrapositive technique. As for what you have in your question, as one of the comments state, I'm not sure how you get that $3k + 1 = 3n$, i.e., where does the "$3$" part come from in $3n$?
          $endgroup$
          – John Omielan
          Jan 20 at 2:17
















        0












        $begingroup$

        I believe a simpler way is to use the unique factorization to show that for



        $$n = prod_{i , = , 1}^{m} p_i^{a_i} tag{1}label{eq1}$$



        where the $p_i$ are unique primes, means that



        $$n^2 = prod_{i , = , 1}^{m} p_i^{2a_i} tag{2}label{eq2}$$



        Thus, if $3 ; vert ; n^2$, then one of the $p_i$ must be $3$, so $3 ; vert ; n$.



        Also, using your suggestion, if $n = 3k + 1$ or $n = 3k + 2$, then $n^2 = 9k^2 + 6k + 1$ or $n^2 = 9k^2 + 12k + 4$. In either case, when dividing by $3$, there is a remainder of $1$, showing that $3$ doesn't divide $n^2$. But, as $3 ; vert ; n^2$, then $n$ cannot be either of the $2$ forms, so it must be $n = 3k$, giving that $3 ; vert ; n$. This is an example of proving what's requested using contrapositive as it shows that if $3 not{vert} ; n$, then $3 not{vert} ; n^2$.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Thanks for you comment. We have not gotten to unique factorization yet in mathematical reasoning class. Would my proof suffice?
          $endgroup$
          – John
          Jan 20 at 2:00










        • $begingroup$
          @John I was editing my response when you wrote your comment. I hope it sufficiently answers your question.
          $endgroup$
          – John Omielan
          Jan 20 at 2:02










        • $begingroup$
          @John Based on the definition for contrapositive, I believe what I showed in the last paragraph uses the contrapositive technique. As for what you have in your question, as one of the comments state, I'm not sure how you get that $3k + 1 = 3n$, i.e., where does the "$3$" part come from in $3n$?
          $endgroup$
          – John Omielan
          Jan 20 at 2:17














        0












        0








        0





        $begingroup$

        I believe a simpler way is to use the unique factorization to show that for



        $$n = prod_{i , = , 1}^{m} p_i^{a_i} tag{1}label{eq1}$$



        where the $p_i$ are unique primes, means that



        $$n^2 = prod_{i , = , 1}^{m} p_i^{2a_i} tag{2}label{eq2}$$



        Thus, if $3 ; vert ; n^2$, then one of the $p_i$ must be $3$, so $3 ; vert ; n$.



        Also, using your suggestion, if $n = 3k + 1$ or $n = 3k + 2$, then $n^2 = 9k^2 + 6k + 1$ or $n^2 = 9k^2 + 12k + 4$. In either case, when dividing by $3$, there is a remainder of $1$, showing that $3$ doesn't divide $n^2$. But, as $3 ; vert ; n^2$, then $n$ cannot be either of the $2$ forms, so it must be $n = 3k$, giving that $3 ; vert ; n$. This is an example of proving what's requested using contrapositive as it shows that if $3 not{vert} ; n$, then $3 not{vert} ; n^2$.






        share|cite|improve this answer











        $endgroup$



        I believe a simpler way is to use the unique factorization to show that for



        $$n = prod_{i , = , 1}^{m} p_i^{a_i} tag{1}label{eq1}$$



        where the $p_i$ are unique primes, means that



        $$n^2 = prod_{i , = , 1}^{m} p_i^{2a_i} tag{2}label{eq2}$$



        Thus, if $3 ; vert ; n^2$, then one of the $p_i$ must be $3$, so $3 ; vert ; n$.



        Also, using your suggestion, if $n = 3k + 1$ or $n = 3k + 2$, then $n^2 = 9k^2 + 6k + 1$ or $n^2 = 9k^2 + 12k + 4$. In either case, when dividing by $3$, there is a remainder of $1$, showing that $3$ doesn't divide $n^2$. But, as $3 ; vert ; n^2$, then $n$ cannot be either of the $2$ forms, so it must be $n = 3k$, giving that $3 ; vert ; n$. This is an example of proving what's requested using contrapositive as it shows that if $3 not{vert} ; n$, then $3 not{vert} ; n^2$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 20 at 2:15

























        answered Jan 20 at 1:59









        John OmielanJohn Omielan

        2,849212




        2,849212












        • $begingroup$
          Thanks for you comment. We have not gotten to unique factorization yet in mathematical reasoning class. Would my proof suffice?
          $endgroup$
          – John
          Jan 20 at 2:00










        • $begingroup$
          @John I was editing my response when you wrote your comment. I hope it sufficiently answers your question.
          $endgroup$
          – John Omielan
          Jan 20 at 2:02










        • $begingroup$
          @John Based on the definition for contrapositive, I believe what I showed in the last paragraph uses the contrapositive technique. As for what you have in your question, as one of the comments state, I'm not sure how you get that $3k + 1 = 3n$, i.e., where does the "$3$" part come from in $3n$?
          $endgroup$
          – John Omielan
          Jan 20 at 2:17


















        • $begingroup$
          Thanks for you comment. We have not gotten to unique factorization yet in mathematical reasoning class. Would my proof suffice?
          $endgroup$
          – John
          Jan 20 at 2:00










        • $begingroup$
          @John I was editing my response when you wrote your comment. I hope it sufficiently answers your question.
          $endgroup$
          – John Omielan
          Jan 20 at 2:02










        • $begingroup$
          @John Based on the definition for contrapositive, I believe what I showed in the last paragraph uses the contrapositive technique. As for what you have in your question, as one of the comments state, I'm not sure how you get that $3k + 1 = 3n$, i.e., where does the "$3$" part come from in $3n$?
          $endgroup$
          – John Omielan
          Jan 20 at 2:17
















        $begingroup$
        Thanks for you comment. We have not gotten to unique factorization yet in mathematical reasoning class. Would my proof suffice?
        $endgroup$
        – John
        Jan 20 at 2:00




        $begingroup$
        Thanks for you comment. We have not gotten to unique factorization yet in mathematical reasoning class. Would my proof suffice?
        $endgroup$
        – John
        Jan 20 at 2:00












        $begingroup$
        @John I was editing my response when you wrote your comment. I hope it sufficiently answers your question.
        $endgroup$
        – John Omielan
        Jan 20 at 2:02




        $begingroup$
        @John I was editing my response when you wrote your comment. I hope it sufficiently answers your question.
        $endgroup$
        – John Omielan
        Jan 20 at 2:02












        $begingroup$
        @John Based on the definition for contrapositive, I believe what I showed in the last paragraph uses the contrapositive technique. As for what you have in your question, as one of the comments state, I'm not sure how you get that $3k + 1 = 3n$, i.e., where does the "$3$" part come from in $3n$?
        $endgroup$
        – John Omielan
        Jan 20 at 2:17




        $begingroup$
        @John Based on the definition for contrapositive, I believe what I showed in the last paragraph uses the contrapositive technique. As for what you have in your question, as one of the comments state, I'm not sure how you get that $3k + 1 = 3n$, i.e., where does the "$3$" part come from in $3n$?
        $endgroup$
        – John Omielan
        Jan 20 at 2:17











        0












        $begingroup$

        Presumably you have had the division theorem.



        For $n$ and integer there exist integer $k, r$ so that $n = 3k + r$ where $0 le r < 3$. So $r = 0, 1$ or $2$.



        Can you accept that?



        If $r=0$ then $3|n$.



        If $r = 1$ then $n^2 = (3k + 1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1$ and $3not mid n^2$ and that's a contradiction.



        If $r = 2$ then $n^2= (3k + 2)^2 = 9k^2 + 12k + 4 = 3(k^2 + 4k+ 1) + 1$ and $3not mid n^2$ and that's a contradiction.



        So if $3|n^2$ then the only possibility is $n = 3k$ for some $k$.






        share|cite|improve this answer









        $endgroup$


















          0












          $begingroup$

          Presumably you have had the division theorem.



          For $n$ and integer there exist integer $k, r$ so that $n = 3k + r$ where $0 le r < 3$. So $r = 0, 1$ or $2$.



          Can you accept that?



          If $r=0$ then $3|n$.



          If $r = 1$ then $n^2 = (3k + 1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1$ and $3not mid n^2$ and that's a contradiction.



          If $r = 2$ then $n^2= (3k + 2)^2 = 9k^2 + 12k + 4 = 3(k^2 + 4k+ 1) + 1$ and $3not mid n^2$ and that's a contradiction.



          So if $3|n^2$ then the only possibility is $n = 3k$ for some $k$.






          share|cite|improve this answer









          $endgroup$
















            0












            0








            0





            $begingroup$

            Presumably you have had the division theorem.



            For $n$ and integer there exist integer $k, r$ so that $n = 3k + r$ where $0 le r < 3$. So $r = 0, 1$ or $2$.



            Can you accept that?



            If $r=0$ then $3|n$.



            If $r = 1$ then $n^2 = (3k + 1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1$ and $3not mid n^2$ and that's a contradiction.



            If $r = 2$ then $n^2= (3k + 2)^2 = 9k^2 + 12k + 4 = 3(k^2 + 4k+ 1) + 1$ and $3not mid n^2$ and that's a contradiction.



            So if $3|n^2$ then the only possibility is $n = 3k$ for some $k$.






            share|cite|improve this answer









            $endgroup$



            Presumably you have had the division theorem.



            For $n$ and integer there exist integer $k, r$ so that $n = 3k + r$ where $0 le r < 3$. So $r = 0, 1$ or $2$.



            Can you accept that?



            If $r=0$ then $3|n$.



            If $r = 1$ then $n^2 = (3k + 1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1$ and $3not mid n^2$ and that's a contradiction.



            If $r = 2$ then $n^2= (3k + 2)^2 = 9k^2 + 12k + 4 = 3(k^2 + 4k+ 1) + 1$ and $3not mid n^2$ and that's a contradiction.



            So if $3|n^2$ then the only possibility is $n = 3k$ for some $k$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 20 at 2:17









            fleabloodfleablood

            71k22686




            71k22686























                0












                $begingroup$

                You don't need to resort to proving the contrapositive; it's possible to prove the statement directly:



                If we take as given that $3$ divides (exactly) one of the three consecutive numbers $n-1$, $n$, and $n+1$, then $3$ divides their product, $(n-1)n(n+1)=n^3-n$. Now if $3$ divides $n^2$, then it also divides $n^3$, and thus it divides the difference, $n^3-(n^3-n)=n$.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  You don't need to resort to proving the contrapositive; it's possible to prove the statement directly:



                  If we take as given that $3$ divides (exactly) one of the three consecutive numbers $n-1$, $n$, and $n+1$, then $3$ divides their product, $(n-1)n(n+1)=n^3-n$. Now if $3$ divides $n^2$, then it also divides $n^3$, and thus it divides the difference, $n^3-(n^3-n)=n$.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    You don't need to resort to proving the contrapositive; it's possible to prove the statement directly:



                    If we take as given that $3$ divides (exactly) one of the three consecutive numbers $n-1$, $n$, and $n+1$, then $3$ divides their product, $(n-1)n(n+1)=n^3-n$. Now if $3$ divides $n^2$, then it also divides $n^3$, and thus it divides the difference, $n^3-(n^3-n)=n$.






                    share|cite|improve this answer









                    $endgroup$



                    You don't need to resort to proving the contrapositive; it's possible to prove the statement directly:



                    If we take as given that $3$ divides (exactly) one of the three consecutive numbers $n-1$, $n$, and $n+1$, then $3$ divides their product, $(n-1)n(n+1)=n^3-n$. Now if $3$ divides $n^2$, then it also divides $n^3$, and thus it divides the difference, $n^3-(n^3-n)=n$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 20 at 2:47









                    Barry CipraBarry Cipra

                    59.6k653126




                    59.6k653126























                        0












                        $begingroup$

                        In
                        If $n mid a^2 $, what is the largest $m$ for which $m mid a$?,
                        I prove this result:



                        Given
                        $n = prod p_i^{a_i}$,
                        then the largest $m$
                        such that
                        $m | a$ for all $a$
                        such that
                        $n | a^2$ is
                        $m = prod p_i^{lceil frac{a_i}{2}rceil}$.



                        If $n$ is a prime,
                        $3$ in this problems case,
                        then
                        $n = 3^1$
                        so
                        $m = 3^1$
                        also.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          In
                          If $n mid a^2 $, what is the largest $m$ for which $m mid a$?,
                          I prove this result:



                          Given
                          $n = prod p_i^{a_i}$,
                          then the largest $m$
                          such that
                          $m | a$ for all $a$
                          such that
                          $n | a^2$ is
                          $m = prod p_i^{lceil frac{a_i}{2}rceil}$.



                          If $n$ is a prime,
                          $3$ in this problems case,
                          then
                          $n = 3^1$
                          so
                          $m = 3^1$
                          also.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            In
                            If $n mid a^2 $, what is the largest $m$ for which $m mid a$?,
                            I prove this result:



                            Given
                            $n = prod p_i^{a_i}$,
                            then the largest $m$
                            such that
                            $m | a$ for all $a$
                            such that
                            $n | a^2$ is
                            $m = prod p_i^{lceil frac{a_i}{2}rceil}$.



                            If $n$ is a prime,
                            $3$ in this problems case,
                            then
                            $n = 3^1$
                            so
                            $m = 3^1$
                            also.






                            share|cite|improve this answer









                            $endgroup$



                            In
                            If $n mid a^2 $, what is the largest $m$ for which $m mid a$?,
                            I prove this result:



                            Given
                            $n = prod p_i^{a_i}$,
                            then the largest $m$
                            such that
                            $m | a$ for all $a$
                            such that
                            $n | a^2$ is
                            $m = prod p_i^{lceil frac{a_i}{2}rceil}$.



                            If $n$ is a prime,
                            $3$ in this problems case,
                            then
                            $n = 3^1$
                            so
                            $m = 3^1$
                            also.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 20 at 3:00









                            marty cohenmarty cohen

                            73.7k549128




                            73.7k549128






























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