Assuming $g$ is a primitive root modulo a prime $p$, show that $p-g$ is a primitive root if and only if $p...












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Assume $g$ is a primitive root modulo a prime $p$. Show that $p-g$ is a primitive root if and only if $p equiv 1 pmod 4$.




I am studying for a number theory exam that is why I am posting a lot of questions related to primitive roots. I would really appreciate your help because they seem like really nice questions.






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    $begingroup$



    Assume $g$ is a primitive root modulo a prime $p$. Show that $p-g$ is a primitive root if and only if $p equiv 1 pmod 4$.




    I am studying for a number theory exam that is why I am posting a lot of questions related to primitive roots. I would really appreciate your help because they seem like really nice questions.






    elementary-number-theory







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$



      Assume $g$ is a primitive root modulo a prime $p$. Show that $p-g$ is a primitive root if and only if $p equiv 1 pmod 4$.




      I am studying for a number theory exam that is why I am posting a lot of questions related to primitive roots. I would really appreciate your help because they seem like really nice questions.






      elementary-number-theory







      share|cite|improve this question











      $endgroup$





      Assume $g$ is a primitive root modulo a prime $p$. Show that $p-g$ is a primitive root if and only if $p equiv 1 pmod 4$.




      I am studying for a number theory exam that is why I am posting a lot of questions related to primitive roots. I would really appreciate your help because they seem like really nice questions.






      elementary-number-theory




      elementary-number-theory






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      edited Apr 16 '15 at 4:35









      kate

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      asked Apr 10 '15 at 22:16









      user2214user2214

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          $begingroup$

          We solve the first problem. The case $p=2$ is simple, so let $p$ be an odd prime.



          If $pequiv 3pmod{4}$, then $p-g$ is not a primitive root of $p$. For $p-gequiv -gpmod{p}$. But $-1$ is a quadratic non-residue of $p$, and therefore $-g$ is a QR of $p$, so cannot be a primitive root of $p$.



          Conversely, let $pequiv 1pmod{4}$. We show that $-g$ is a primitive root of $p$.



          By Fermat's Theorem, $gequiv g^pequiv -(-g)^ppmod{p}$. But $-1$ is a QR of $p$, so $-1equiv g^{2k}equiv (-g)^{2k}pmof{p}$ for some integer $k$.



          t follows that $gequiv (-g)^{2k}(-g)^ppmod{p}$. Thus $g$ is congruent to a power of $-g$. Since the powers of $g$ travel, modulo $p$, through $1,2,dots, p-1$, so do the powers of $-g$, and therefore $-g$ is a primitive root of $p$.



          Remark: For other problems that you may post (and this one also) please indicate what you have tried, any progress you may have made, and where difficulties remain.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for solving the first question, but I believe I need some help for the 2nd one as well which seems more complicated.So since g is a primitive root mod p that we know that $g^{p-1}=1$ (mod p). We also know that if g is a primitive root then g or g+p is a primitive root of $p^2$.But How do I relate this facts to the 2nd question? @AndreNicolas
            $endgroup$
            – user2214
            Apr 11 '15 at 20:13








          • 1




            $begingroup$
            Your are welcome. Multipart questions in which the parts are not closely related often get closed quickly. I would prefer two questions, the second about primitive roots mod $p^2$. Recall that $g+kp$ is a primitive root mod $p^2$ for all but one $k$ in the interval $0$ to $p-1$.
            $endgroup$
            – André Nicolas
            Apr 11 '15 at 22:34



















          1












          $begingroup$

          $$-g=g(-1)equiv g^{1+(p-1)/2}equiv g^{(p+1)/2}$$



          We know, ord$_ma=d, $ord$_m(a^k)=dfrac d{(d,k)}$ (Proof @Page$#95$)



          ord$_p g^{(p+1)/2}=dfrac{p-1}{left(p-1,dfrac{p+1}2right)}$



          Now, if integer $d(>0)$ divides both, $d$ must divide $-(p-1)+2cdotdfrac{p+1}2=2$



          As for odd prime $p,p-1$ is even



          $left(p-1,dfrac{p+1}2right)=2$ if $dfrac{p+1}2$ is even $=2k$(say) $iff p=4k-1equiv-1pmod4$



          $left(p-1,dfrac{p+1}2right)=1$ if $dfrac{p+1}2$ is odd $=2k+1$(say) $iff p=4k+1equiv1pmod4$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            See also: math.stackexchange.com/questions/56028/…
            $endgroup$
            – lab bhattacharjee
            Dec 6 '18 at 11:44











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          2 Answers
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          3












          $begingroup$

          We solve the first problem. The case $p=2$ is simple, so let $p$ be an odd prime.



          If $pequiv 3pmod{4}$, then $p-g$ is not a primitive root of $p$. For $p-gequiv -gpmod{p}$. But $-1$ is a quadratic non-residue of $p$, and therefore $-g$ is a QR of $p$, so cannot be a primitive root of $p$.



          Conversely, let $pequiv 1pmod{4}$. We show that $-g$ is a primitive root of $p$.



          By Fermat's Theorem, $gequiv g^pequiv -(-g)^ppmod{p}$. But $-1$ is a QR of $p$, so $-1equiv g^{2k}equiv (-g)^{2k}pmof{p}$ for some integer $k$.



          t follows that $gequiv (-g)^{2k}(-g)^ppmod{p}$. Thus $g$ is congruent to a power of $-g$. Since the powers of $g$ travel, modulo $p$, through $1,2,dots, p-1$, so do the powers of $-g$, and therefore $-g$ is a primitive root of $p$.



          Remark: For other problems that you may post (and this one also) please indicate what you have tried, any progress you may have made, and where difficulties remain.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for solving the first question, but I believe I need some help for the 2nd one as well which seems more complicated.So since g is a primitive root mod p that we know that $g^{p-1}=1$ (mod p). We also know that if g is a primitive root then g or g+p is a primitive root of $p^2$.But How do I relate this facts to the 2nd question? @AndreNicolas
            $endgroup$
            – user2214
            Apr 11 '15 at 20:13








          • 1




            $begingroup$
            Your are welcome. Multipart questions in which the parts are not closely related often get closed quickly. I would prefer two questions, the second about primitive roots mod $p^2$. Recall that $g+kp$ is a primitive root mod $p^2$ for all but one $k$ in the interval $0$ to $p-1$.
            $endgroup$
            – André Nicolas
            Apr 11 '15 at 22:34
















          3












          $begingroup$

          We solve the first problem. The case $p=2$ is simple, so let $p$ be an odd prime.



          If $pequiv 3pmod{4}$, then $p-g$ is not a primitive root of $p$. For $p-gequiv -gpmod{p}$. But $-1$ is a quadratic non-residue of $p$, and therefore $-g$ is a QR of $p$, so cannot be a primitive root of $p$.



          Conversely, let $pequiv 1pmod{4}$. We show that $-g$ is a primitive root of $p$.



          By Fermat's Theorem, $gequiv g^pequiv -(-g)^ppmod{p}$. But $-1$ is a QR of $p$, so $-1equiv g^{2k}equiv (-g)^{2k}pmof{p}$ for some integer $k$.



          t follows that $gequiv (-g)^{2k}(-g)^ppmod{p}$. Thus $g$ is congruent to a power of $-g$. Since the powers of $g$ travel, modulo $p$, through $1,2,dots, p-1$, so do the powers of $-g$, and therefore $-g$ is a primitive root of $p$.



          Remark: For other problems that you may post (and this one also) please indicate what you have tried, any progress you may have made, and where difficulties remain.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for solving the first question, but I believe I need some help for the 2nd one as well which seems more complicated.So since g is a primitive root mod p that we know that $g^{p-1}=1$ (mod p). We also know that if g is a primitive root then g or g+p is a primitive root of $p^2$.But How do I relate this facts to the 2nd question? @AndreNicolas
            $endgroup$
            – user2214
            Apr 11 '15 at 20:13








          • 1




            $begingroup$
            Your are welcome. Multipart questions in which the parts are not closely related often get closed quickly. I would prefer two questions, the second about primitive roots mod $p^2$. Recall that $g+kp$ is a primitive root mod $p^2$ for all but one $k$ in the interval $0$ to $p-1$.
            $endgroup$
            – André Nicolas
            Apr 11 '15 at 22:34














          3












          3








          3





          $begingroup$

          We solve the first problem. The case $p=2$ is simple, so let $p$ be an odd prime.



          If $pequiv 3pmod{4}$, then $p-g$ is not a primitive root of $p$. For $p-gequiv -gpmod{p}$. But $-1$ is a quadratic non-residue of $p$, and therefore $-g$ is a QR of $p$, so cannot be a primitive root of $p$.



          Conversely, let $pequiv 1pmod{4}$. We show that $-g$ is a primitive root of $p$.



          By Fermat's Theorem, $gequiv g^pequiv -(-g)^ppmod{p}$. But $-1$ is a QR of $p$, so $-1equiv g^{2k}equiv (-g)^{2k}pmof{p}$ for some integer $k$.



          t follows that $gequiv (-g)^{2k}(-g)^ppmod{p}$. Thus $g$ is congruent to a power of $-g$. Since the powers of $g$ travel, modulo $p$, through $1,2,dots, p-1$, so do the powers of $-g$, and therefore $-g$ is a primitive root of $p$.



          Remark: For other problems that you may post (and this one also) please indicate what you have tried, any progress you may have made, and where difficulties remain.






          share|cite|improve this answer











          $endgroup$



          We solve the first problem. The case $p=2$ is simple, so let $p$ be an odd prime.



          If $pequiv 3pmod{4}$, then $p-g$ is not a primitive root of $p$. For $p-gequiv -gpmod{p}$. But $-1$ is a quadratic non-residue of $p$, and therefore $-g$ is a QR of $p$, so cannot be a primitive root of $p$.



          Conversely, let $pequiv 1pmod{4}$. We show that $-g$ is a primitive root of $p$.



          By Fermat's Theorem, $gequiv g^pequiv -(-g)^ppmod{p}$. But $-1$ is a QR of $p$, so $-1equiv g^{2k}equiv (-g)^{2k}pmof{p}$ for some integer $k$.



          t follows that $gequiv (-g)^{2k}(-g)^ppmod{p}$. Thus $g$ is congruent to a power of $-g$. Since the powers of $g$ travel, modulo $p$, through $1,2,dots, p-1$, so do the powers of $-g$, and therefore $-g$ is a primitive root of $p$.



          Remark: For other problems that you may post (and this one also) please indicate what you have tried, any progress you may have made, and where difficulties remain.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 10 '15 at 22:47

























          answered Apr 10 '15 at 22:40









          André NicolasAndré Nicolas

          452k36425810




          452k36425810












          • $begingroup$
            Thank you for solving the first question, but I believe I need some help for the 2nd one as well which seems more complicated.So since g is a primitive root mod p that we know that $g^{p-1}=1$ (mod p). We also know that if g is a primitive root then g or g+p is a primitive root of $p^2$.But How do I relate this facts to the 2nd question? @AndreNicolas
            $endgroup$
            – user2214
            Apr 11 '15 at 20:13








          • 1




            $begingroup$
            Your are welcome. Multipart questions in which the parts are not closely related often get closed quickly. I would prefer two questions, the second about primitive roots mod $p^2$. Recall that $g+kp$ is a primitive root mod $p^2$ for all but one $k$ in the interval $0$ to $p-1$.
            $endgroup$
            – André Nicolas
            Apr 11 '15 at 22:34


















          • $begingroup$
            Thank you for solving the first question, but I believe I need some help for the 2nd one as well which seems more complicated.So since g is a primitive root mod p that we know that $g^{p-1}=1$ (mod p). We also know that if g is a primitive root then g or g+p is a primitive root of $p^2$.But How do I relate this facts to the 2nd question? @AndreNicolas
            $endgroup$
            – user2214
            Apr 11 '15 at 20:13








          • 1




            $begingroup$
            Your are welcome. Multipart questions in which the parts are not closely related often get closed quickly. I would prefer two questions, the second about primitive roots mod $p^2$. Recall that $g+kp$ is a primitive root mod $p^2$ for all but one $k$ in the interval $0$ to $p-1$.
            $endgroup$
            – André Nicolas
            Apr 11 '15 at 22:34
















          $begingroup$
          Thank you for solving the first question, but I believe I need some help for the 2nd one as well which seems more complicated.So since g is a primitive root mod p that we know that $g^{p-1}=1$ (mod p). We also know that if g is a primitive root then g or g+p is a primitive root of $p^2$.But How do I relate this facts to the 2nd question? @AndreNicolas
          $endgroup$
          – user2214
          Apr 11 '15 at 20:13






          $begingroup$
          Thank you for solving the first question, but I believe I need some help for the 2nd one as well which seems more complicated.So since g is a primitive root mod p that we know that $g^{p-1}=1$ (mod p). We also know that if g is a primitive root then g or g+p is a primitive root of $p^2$.But How do I relate this facts to the 2nd question? @AndreNicolas
          $endgroup$
          – user2214
          Apr 11 '15 at 20:13






          1




          1




          $begingroup$
          Your are welcome. Multipart questions in which the parts are not closely related often get closed quickly. I would prefer two questions, the second about primitive roots mod $p^2$. Recall that $g+kp$ is a primitive root mod $p^2$ for all but one $k$ in the interval $0$ to $p-1$.
          $endgroup$
          – André Nicolas
          Apr 11 '15 at 22:34




          $begingroup$
          Your are welcome. Multipart questions in which the parts are not closely related often get closed quickly. I would prefer two questions, the second about primitive roots mod $p^2$. Recall that $g+kp$ is a primitive root mod $p^2$ for all but one $k$ in the interval $0$ to $p-1$.
          $endgroup$
          – André Nicolas
          Apr 11 '15 at 22:34











          1












          $begingroup$

          $$-g=g(-1)equiv g^{1+(p-1)/2}equiv g^{(p+1)/2}$$



          We know, ord$_ma=d, $ord$_m(a^k)=dfrac d{(d,k)}$ (Proof @Page$#95$)



          ord$_p g^{(p+1)/2}=dfrac{p-1}{left(p-1,dfrac{p+1}2right)}$



          Now, if integer $d(>0)$ divides both, $d$ must divide $-(p-1)+2cdotdfrac{p+1}2=2$



          As for odd prime $p,p-1$ is even



          $left(p-1,dfrac{p+1}2right)=2$ if $dfrac{p+1}2$ is even $=2k$(say) $iff p=4k-1equiv-1pmod4$



          $left(p-1,dfrac{p+1}2right)=1$ if $dfrac{p+1}2$ is odd $=2k+1$(say) $iff p=4k+1equiv1pmod4$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            See also: math.stackexchange.com/questions/56028/…
            $endgroup$
            – lab bhattacharjee
            Dec 6 '18 at 11:44
















          1












          $begingroup$

          $$-g=g(-1)equiv g^{1+(p-1)/2}equiv g^{(p+1)/2}$$



          We know, ord$_ma=d, $ord$_m(a^k)=dfrac d{(d,k)}$ (Proof @Page$#95$)



          ord$_p g^{(p+1)/2}=dfrac{p-1}{left(p-1,dfrac{p+1}2right)}$



          Now, if integer $d(>0)$ divides both, $d$ must divide $-(p-1)+2cdotdfrac{p+1}2=2$



          As for odd prime $p,p-1$ is even



          $left(p-1,dfrac{p+1}2right)=2$ if $dfrac{p+1}2$ is even $=2k$(say) $iff p=4k-1equiv-1pmod4$



          $left(p-1,dfrac{p+1}2right)=1$ if $dfrac{p+1}2$ is odd $=2k+1$(say) $iff p=4k+1equiv1pmod4$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            See also: math.stackexchange.com/questions/56028/…
            $endgroup$
            – lab bhattacharjee
            Dec 6 '18 at 11:44














          1












          1








          1





          $begingroup$

          $$-g=g(-1)equiv g^{1+(p-1)/2}equiv g^{(p+1)/2}$$



          We know, ord$_ma=d, $ord$_m(a^k)=dfrac d{(d,k)}$ (Proof @Page$#95$)



          ord$_p g^{(p+1)/2}=dfrac{p-1}{left(p-1,dfrac{p+1}2right)}$



          Now, if integer $d(>0)$ divides both, $d$ must divide $-(p-1)+2cdotdfrac{p+1}2=2$



          As for odd prime $p,p-1$ is even



          $left(p-1,dfrac{p+1}2right)=2$ if $dfrac{p+1}2$ is even $=2k$(say) $iff p=4k-1equiv-1pmod4$



          $left(p-1,dfrac{p+1}2right)=1$ if $dfrac{p+1}2$ is odd $=2k+1$(say) $iff p=4k+1equiv1pmod4$






          share|cite|improve this answer









          $endgroup$



          $$-g=g(-1)equiv g^{1+(p-1)/2}equiv g^{(p+1)/2}$$



          We know, ord$_ma=d, $ord$_m(a^k)=dfrac d{(d,k)}$ (Proof @Page$#95$)



          ord$_p g^{(p+1)/2}=dfrac{p-1}{left(p-1,dfrac{p+1}2right)}$



          Now, if integer $d(>0)$ divides both, $d$ must divide $-(p-1)+2cdotdfrac{p+1}2=2$



          As for odd prime $p,p-1$ is even



          $left(p-1,dfrac{p+1}2right)=2$ if $dfrac{p+1}2$ is even $=2k$(say) $iff p=4k-1equiv-1pmod4$



          $left(p-1,dfrac{p+1}2right)=1$ if $dfrac{p+1}2$ is odd $=2k+1$(say) $iff p=4k+1equiv1pmod4$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 11:40









          lab bhattacharjeelab bhattacharjee

          225k15157275




          225k15157275












          • $begingroup$
            See also: math.stackexchange.com/questions/56028/…
            $endgroup$
            – lab bhattacharjee
            Dec 6 '18 at 11:44


















          • $begingroup$
            See also: math.stackexchange.com/questions/56028/…
            $endgroup$
            – lab bhattacharjee
            Dec 6 '18 at 11:44
















          $begingroup$
          See also: math.stackexchange.com/questions/56028/…
          $endgroup$
          – lab bhattacharjee
          Dec 6 '18 at 11:44




          $begingroup$
          See also: math.stackexchange.com/questions/56028/…
          $endgroup$
          – lab bhattacharjee
          Dec 6 '18 at 11:44


















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          Ida-Boy-Ed-Garten

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          Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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