For any integer $a$, $gcd(11a+5,2a+1)=1$.












1












$begingroup$


How would I go by proving this statement?



What I did was I tried using Proposition GCD Of One, so that



$(11a+5)x + (2a+1)y = 1$, and

$(11x+2y)a + (5x+y) = 1$.



But I have no idea what to do from here.
What can I do?






divisibility







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    How would I go by proving this statement?



    What I did was I tried using Proposition GCD Of One, so that



    $(11a+5)x + (2a+1)y = 1$, and

    $(11x+2y)a + (5x+y) = 1$.



    But I have no idea what to do from here.
    What can I do?






    divisibility







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      How would I go by proving this statement?



      What I did was I tried using Proposition GCD Of One, so that



      $(11a+5)x + (2a+1)y = 1$, and

      $(11x+2y)a + (5x+y) = 1$.



      But I have no idea what to do from here.
      What can I do?






      divisibility







      share|cite|improve this question









      $endgroup$




      How would I go by proving this statement?



      What I did was I tried using Proposition GCD Of One, so that



      $(11a+5)x + (2a+1)y = 1$, and

      $(11x+2y)a + (5x+y) = 1$.



      But I have no idea what to do from here.
      What can I do?






      divisibility




      divisibility






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Apr 15 '14 at 0:29









      HaxifyHaxify

      150116




      150116






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          Hint: Can you justify the following?



          begin{align*}
          gcd(11a + 5, 2a + 1) &= gcdBig((11a + 5) - 5(2a + 1), 2a + 1Big) \
          &= gcd(a, 2a + 1)
          end{align*}



          Can you finish it from here?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Sorry, but could you explain how the first line works?
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:33










          • $begingroup$
            @Haxify $$gcd(a, c) = gcd(a - c, c)$$
            $endgroup$
            – user61527
            Apr 15 '14 at 0:35










          • $begingroup$
            Oh, okay I see how that works. Thank you!
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:37










          • $begingroup$
            Then is it correct to say that $gcd(11a+5,2a+1)=gcd(a,2a+1)=gcd(1,a)$?
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:47












          • $begingroup$
            @Haxify Exactly.
            $endgroup$
            – user61527
            Apr 15 '14 at 0:47



















          2












          $begingroup$

          Hint $ $ If $,d,$ is a common factor then $,{rm mod} d!: 2a+1equiv 0equiv 11a+5.,$ Now eliminate $,a,$ by cross multiplying $ 0equiv color{#0a0}{11}(2a+1)-color{#c00}2(11a+5) equiv color{#0a0}{11}-color{#c00}{10}equiv 1 $ i.e. $,0equiv 1,,$ so $, dmid 1!-!0,,$ so $,d = 1.$



          Remark $ $ More conceptually, put the fractions for $,-a,$ over the common denominator $,22$



          $$ {rm mod} d!: frac{1}{2} equiv -a,equiv frac{5}{11} Rightarrow frac{color{#0a0}{11}}{22},equiv, frac{color{#c00}{10}}{22}qquad$$



          Note $ $ The fractions uniquely exist mod $,d,$ since $,2,11,$ are invertible, being coprime to $,d,$ (else $2mid dmid 2a+1,Rightarrow 2mid 1,,$ and $,11mid dmid 11a+5,Rightarrow, 11mid 5,)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh, I see. Thank you!!
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:38












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          2 Answers
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          active

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          2 Answers
          2






          active

          oldest

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          active

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          active

          oldest

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          4












          $begingroup$

          Hint: Can you justify the following?



          begin{align*}
          gcd(11a + 5, 2a + 1) &= gcdBig((11a + 5) - 5(2a + 1), 2a + 1Big) \
          &= gcd(a, 2a + 1)
          end{align*}



          Can you finish it from here?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Sorry, but could you explain how the first line works?
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:33










          • $begingroup$
            @Haxify $$gcd(a, c) = gcd(a - c, c)$$
            $endgroup$
            – user61527
            Apr 15 '14 at 0:35










          • $begingroup$
            Oh, okay I see how that works. Thank you!
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:37










          • $begingroup$
            Then is it correct to say that $gcd(11a+5,2a+1)=gcd(a,2a+1)=gcd(1,a)$?
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:47












          • $begingroup$
            @Haxify Exactly.
            $endgroup$
            – user61527
            Apr 15 '14 at 0:47
















          4












          $begingroup$

          Hint: Can you justify the following?



          begin{align*}
          gcd(11a + 5, 2a + 1) &= gcdBig((11a + 5) - 5(2a + 1), 2a + 1Big) \
          &= gcd(a, 2a + 1)
          end{align*}



          Can you finish it from here?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Sorry, but could you explain how the first line works?
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:33










          • $begingroup$
            @Haxify $$gcd(a, c) = gcd(a - c, c)$$
            $endgroup$
            – user61527
            Apr 15 '14 at 0:35










          • $begingroup$
            Oh, okay I see how that works. Thank you!
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:37










          • $begingroup$
            Then is it correct to say that $gcd(11a+5,2a+1)=gcd(a,2a+1)=gcd(1,a)$?
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:47












          • $begingroup$
            @Haxify Exactly.
            $endgroup$
            – user61527
            Apr 15 '14 at 0:47














          4












          4








          4





          $begingroup$

          Hint: Can you justify the following?



          begin{align*}
          gcd(11a + 5, 2a + 1) &= gcdBig((11a + 5) - 5(2a + 1), 2a + 1Big) \
          &= gcd(a, 2a + 1)
          end{align*}



          Can you finish it from here?






          share|cite|improve this answer









          $endgroup$



          Hint: Can you justify the following?



          begin{align*}
          gcd(11a + 5, 2a + 1) &= gcdBig((11a + 5) - 5(2a + 1), 2a + 1Big) \
          &= gcd(a, 2a + 1)
          end{align*}



          Can you finish it from here?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 15 '14 at 0:30







          user61527



















          • $begingroup$
            Sorry, but could you explain how the first line works?
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:33










          • $begingroup$
            @Haxify $$gcd(a, c) = gcd(a - c, c)$$
            $endgroup$
            – user61527
            Apr 15 '14 at 0:35










          • $begingroup$
            Oh, okay I see how that works. Thank you!
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:37










          • $begingroup$
            Then is it correct to say that $gcd(11a+5,2a+1)=gcd(a,2a+1)=gcd(1,a)$?
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:47












          • $begingroup$
            @Haxify Exactly.
            $endgroup$
            – user61527
            Apr 15 '14 at 0:47


















          • $begingroup$
            Sorry, but could you explain how the first line works?
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:33










          • $begingroup$
            @Haxify $$gcd(a, c) = gcd(a - c, c)$$
            $endgroup$
            – user61527
            Apr 15 '14 at 0:35










          • $begingroup$
            Oh, okay I see how that works. Thank you!
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:37










          • $begingroup$
            Then is it correct to say that $gcd(11a+5,2a+1)=gcd(a,2a+1)=gcd(1,a)$?
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:47












          • $begingroup$
            @Haxify Exactly.
            $endgroup$
            – user61527
            Apr 15 '14 at 0:47
















          $begingroup$
          Sorry, but could you explain how the first line works?
          $endgroup$
          – Haxify
          Apr 15 '14 at 0:33




          $begingroup$
          Sorry, but could you explain how the first line works?
          $endgroup$
          – Haxify
          Apr 15 '14 at 0:33












          $begingroup$
          @Haxify $$gcd(a, c) = gcd(a - c, c)$$
          $endgroup$
          – user61527
          Apr 15 '14 at 0:35




          $begingroup$
          @Haxify $$gcd(a, c) = gcd(a - c, c)$$
          $endgroup$
          – user61527
          Apr 15 '14 at 0:35












          $begingroup$
          Oh, okay I see how that works. Thank you!
          $endgroup$
          – Haxify
          Apr 15 '14 at 0:37




          $begingroup$
          Oh, okay I see how that works. Thank you!
          $endgroup$
          – Haxify
          Apr 15 '14 at 0:37












          $begingroup$
          Then is it correct to say that $gcd(11a+5,2a+1)=gcd(a,2a+1)=gcd(1,a)$?
          $endgroup$
          – Haxify
          Apr 15 '14 at 0:47






          $begingroup$
          Then is it correct to say that $gcd(11a+5,2a+1)=gcd(a,2a+1)=gcd(1,a)$?
          $endgroup$
          – Haxify
          Apr 15 '14 at 0:47














          $begingroup$
          @Haxify Exactly.
          $endgroup$
          – user61527
          Apr 15 '14 at 0:47




          $begingroup$
          @Haxify Exactly.
          $endgroup$
          – user61527
          Apr 15 '14 at 0:47











          2












          $begingroup$

          Hint $ $ If $,d,$ is a common factor then $,{rm mod} d!: 2a+1equiv 0equiv 11a+5.,$ Now eliminate $,a,$ by cross multiplying $ 0equiv color{#0a0}{11}(2a+1)-color{#c00}2(11a+5) equiv color{#0a0}{11}-color{#c00}{10}equiv 1 $ i.e. $,0equiv 1,,$ so $, dmid 1!-!0,,$ so $,d = 1.$



          Remark $ $ More conceptually, put the fractions for $,-a,$ over the common denominator $,22$



          $$ {rm mod} d!: frac{1}{2} equiv -a,equiv frac{5}{11} Rightarrow frac{color{#0a0}{11}}{22},equiv, frac{color{#c00}{10}}{22}qquad$$



          Note $ $ The fractions uniquely exist mod $,d,$ since $,2,11,$ are invertible, being coprime to $,d,$ (else $2mid dmid 2a+1,Rightarrow 2mid 1,,$ and $,11mid dmid 11a+5,Rightarrow, 11mid 5,)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh, I see. Thank you!!
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:38
















          2












          $begingroup$

          Hint $ $ If $,d,$ is a common factor then $,{rm mod} d!: 2a+1equiv 0equiv 11a+5.,$ Now eliminate $,a,$ by cross multiplying $ 0equiv color{#0a0}{11}(2a+1)-color{#c00}2(11a+5) equiv color{#0a0}{11}-color{#c00}{10}equiv 1 $ i.e. $,0equiv 1,,$ so $, dmid 1!-!0,,$ so $,d = 1.$



          Remark $ $ More conceptually, put the fractions for $,-a,$ over the common denominator $,22$



          $$ {rm mod} d!: frac{1}{2} equiv -a,equiv frac{5}{11} Rightarrow frac{color{#0a0}{11}}{22},equiv, frac{color{#c00}{10}}{22}qquad$$



          Note $ $ The fractions uniquely exist mod $,d,$ since $,2,11,$ are invertible, being coprime to $,d,$ (else $2mid dmid 2a+1,Rightarrow 2mid 1,,$ and $,11mid dmid 11a+5,Rightarrow, 11mid 5,)$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Oh, I see. Thank you!!
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:38














          2












          2








          2





          $begingroup$

          Hint $ $ If $,d,$ is a common factor then $,{rm mod} d!: 2a+1equiv 0equiv 11a+5.,$ Now eliminate $,a,$ by cross multiplying $ 0equiv color{#0a0}{11}(2a+1)-color{#c00}2(11a+5) equiv color{#0a0}{11}-color{#c00}{10}equiv 1 $ i.e. $,0equiv 1,,$ so $, dmid 1!-!0,,$ so $,d = 1.$



          Remark $ $ More conceptually, put the fractions for $,-a,$ over the common denominator $,22$



          $$ {rm mod} d!: frac{1}{2} equiv -a,equiv frac{5}{11} Rightarrow frac{color{#0a0}{11}}{22},equiv, frac{color{#c00}{10}}{22}qquad$$



          Note $ $ The fractions uniquely exist mod $,d,$ since $,2,11,$ are invertible, being coprime to $,d,$ (else $2mid dmid 2a+1,Rightarrow 2mid 1,,$ and $,11mid dmid 11a+5,Rightarrow, 11mid 5,)$.






          share|cite|improve this answer











          $endgroup$



          Hint $ $ If $,d,$ is a common factor then $,{rm mod} d!: 2a+1equiv 0equiv 11a+5.,$ Now eliminate $,a,$ by cross multiplying $ 0equiv color{#0a0}{11}(2a+1)-color{#c00}2(11a+5) equiv color{#0a0}{11}-color{#c00}{10}equiv 1 $ i.e. $,0equiv 1,,$ so $, dmid 1!-!0,,$ so $,d = 1.$



          Remark $ $ More conceptually, put the fractions for $,-a,$ over the common denominator $,22$



          $$ {rm mod} d!: frac{1}{2} equiv -a,equiv frac{5}{11} Rightarrow frac{color{#0a0}{11}}{22},equiv, frac{color{#c00}{10}}{22}qquad$$



          Note $ $ The fractions uniquely exist mod $,d,$ since $,2,11,$ are invertible, being coprime to $,d,$ (else $2mid dmid 2a+1,Rightarrow 2mid 1,,$ and $,11mid dmid 11a+5,Rightarrow, 11mid 5,)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 2 at 4:08

























          answered Apr 15 '14 at 0:35









          Bill DubuqueBill Dubuque

          214k29197656




          214k29197656












          • $begingroup$
            Oh, I see. Thank you!!
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:38


















          • $begingroup$
            Oh, I see. Thank you!!
            $endgroup$
            – Haxify
            Apr 15 '14 at 0:38
















          $begingroup$
          Oh, I see. Thank you!!
          $endgroup$
          – Haxify
          Apr 15 '14 at 0:38




          $begingroup$
          Oh, I see. Thank you!!
          $endgroup$
          – Haxify
          Apr 15 '14 at 0:38


















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