every prime in the form $3k+1$ can be written as the sum of a square and three times a square












3












$begingroup$


i'm having trouble getting started



every prime in the form $3k+1$ can be written as the sum of a square and three times a square. verify this for all appropriate primes less than $100$.



hints would be greatly appreciated!






elementary-number-theory discrete-mathematics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $7$ is a prime of the form $3k+1$, since $7=3 cdot 2 + 1$. Also, $7 = 2^2 + 3 cdot 1^2$. Now, can you do the case for $13$?
    $endgroup$
    – Gunnar Sveinsson
    Jan 25 at 2:38






  • 1




    $begingroup$
    $7 = 4 + 3, ; 13 = 1 + 12, ; 19 = 16 + 3$
    $endgroup$
    – Will Jagy
    Jan 25 at 2:39










  • $begingroup$
    What's the trouble? You don't know what a prime is? You don't know what $3k+1$ means? You don't know what a square is? You don't know what a sum is? You don't know what "verify" means? You don't know what "less than 100" means? Please, give us something to go on, otherwise, how can we help you?
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:01






  • 2




    $begingroup$
    The question, the way you have quoted it, says "verify this for all appropriate primes less than 100," so that's all you have to do. You are being told that it's true for every prime of the form $3k+1$, but you aren't being asked to prove that (which would be considerably harder to do).
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:44






  • 2




    $begingroup$
    Don't. It takes time to learn how to interpret mathematical questions. They're in a foreign language.
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:59
















3












$begingroup$


i'm having trouble getting started



every prime in the form $3k+1$ can be written as the sum of a square and three times a square. verify this for all appropriate primes less than $100$.



hints would be greatly appreciated!






elementary-number-theory discrete-mathematics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $7$ is a prime of the form $3k+1$, since $7=3 cdot 2 + 1$. Also, $7 = 2^2 + 3 cdot 1^2$. Now, can you do the case for $13$?
    $endgroup$
    – Gunnar Sveinsson
    Jan 25 at 2:38






  • 1




    $begingroup$
    $7 = 4 + 3, ; 13 = 1 + 12, ; 19 = 16 + 3$
    $endgroup$
    – Will Jagy
    Jan 25 at 2:39










  • $begingroup$
    What's the trouble? You don't know what a prime is? You don't know what $3k+1$ means? You don't know what a square is? You don't know what a sum is? You don't know what "verify" means? You don't know what "less than 100" means? Please, give us something to go on, otherwise, how can we help you?
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:01






  • 2




    $begingroup$
    The question, the way you have quoted it, says "verify this for all appropriate primes less than 100," so that's all you have to do. You are being told that it's true for every prime of the form $3k+1$, but you aren't being asked to prove that (which would be considerably harder to do).
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:44






  • 2




    $begingroup$
    Don't. It takes time to learn how to interpret mathematical questions. They're in a foreign language.
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:59














3












3








3





$begingroup$


i'm having trouble getting started



every prime in the form $3k+1$ can be written as the sum of a square and three times a square. verify this for all appropriate primes less than $100$.



hints would be greatly appreciated!






elementary-number-theory discrete-mathematics







share|cite|improve this question











$endgroup$




i'm having trouble getting started



every prime in the form $3k+1$ can be written as the sum of a square and three times a square. verify this for all appropriate primes less than $100$.



hints would be greatly appreciated!






elementary-number-theory discrete-mathematics




elementary-number-theory discrete-mathematics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 3 at 17:57









Mauro ALLEGRANZA

66.9k449115




66.9k449115










asked Jan 25 at 2:29









Shana RianShana Rian

243




243








  • 1




    $begingroup$
    $7$ is a prime of the form $3k+1$, since $7=3 cdot 2 + 1$. Also, $7 = 2^2 + 3 cdot 1^2$. Now, can you do the case for $13$?
    $endgroup$
    – Gunnar Sveinsson
    Jan 25 at 2:38






  • 1




    $begingroup$
    $7 = 4 + 3, ; 13 = 1 + 12, ; 19 = 16 + 3$
    $endgroup$
    – Will Jagy
    Jan 25 at 2:39










  • $begingroup$
    What's the trouble? You don't know what a prime is? You don't know what $3k+1$ means? You don't know what a square is? You don't know what a sum is? You don't know what "verify" means? You don't know what "less than 100" means? Please, give us something to go on, otherwise, how can we help you?
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:01






  • 2




    $begingroup$
    The question, the way you have quoted it, says "verify this for all appropriate primes less than 100," so that's all you have to do. You are being told that it's true for every prime of the form $3k+1$, but you aren't being asked to prove that (which would be considerably harder to do).
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:44






  • 2




    $begingroup$
    Don't. It takes time to learn how to interpret mathematical questions. They're in a foreign language.
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:59














  • 1




    $begingroup$
    $7$ is a prime of the form $3k+1$, since $7=3 cdot 2 + 1$. Also, $7 = 2^2 + 3 cdot 1^2$. Now, can you do the case for $13$?
    $endgroup$
    – Gunnar Sveinsson
    Jan 25 at 2:38






  • 1




    $begingroup$
    $7 = 4 + 3, ; 13 = 1 + 12, ; 19 = 16 + 3$
    $endgroup$
    – Will Jagy
    Jan 25 at 2:39










  • $begingroup$
    What's the trouble? You don't know what a prime is? You don't know what $3k+1$ means? You don't know what a square is? You don't know what a sum is? You don't know what "verify" means? You don't know what "less than 100" means? Please, give us something to go on, otherwise, how can we help you?
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:01






  • 2




    $begingroup$
    The question, the way you have quoted it, says "verify this for all appropriate primes less than 100," so that's all you have to do. You are being told that it's true for every prime of the form $3k+1$, but you aren't being asked to prove that (which would be considerably harder to do).
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:44






  • 2




    $begingroup$
    Don't. It takes time to learn how to interpret mathematical questions. They're in a foreign language.
    $endgroup$
    – Gerry Myerson
    Jan 25 at 3:59








1




1




$begingroup$
$7$ is a prime of the form $3k+1$, since $7=3 cdot 2 + 1$. Also, $7 = 2^2 + 3 cdot 1^2$. Now, can you do the case for $13$?
$endgroup$
– Gunnar Sveinsson
Jan 25 at 2:38




$begingroup$
$7$ is a prime of the form $3k+1$, since $7=3 cdot 2 + 1$. Also, $7 = 2^2 + 3 cdot 1^2$. Now, can you do the case for $13$?
$endgroup$
– Gunnar Sveinsson
Jan 25 at 2:38




1




1




$begingroup$
$7 = 4 + 3, ; 13 = 1 + 12, ; 19 = 16 + 3$
$endgroup$
– Will Jagy
Jan 25 at 2:39




$begingroup$
$7 = 4 + 3, ; 13 = 1 + 12, ; 19 = 16 + 3$
$endgroup$
– Will Jagy
Jan 25 at 2:39












$begingroup$
What's the trouble? You don't know what a prime is? You don't know what $3k+1$ means? You don't know what a square is? You don't know what a sum is? You don't know what "verify" means? You don't know what "less than 100" means? Please, give us something to go on, otherwise, how can we help you?
$endgroup$
– Gerry Myerson
Jan 25 at 3:01




$begingroup$
What's the trouble? You don't know what a prime is? You don't know what $3k+1$ means? You don't know what a square is? You don't know what a sum is? You don't know what "verify" means? You don't know what "less than 100" means? Please, give us something to go on, otherwise, how can we help you?
$endgroup$
– Gerry Myerson
Jan 25 at 3:01




2




2




$begingroup$
The question, the way you have quoted it, says "verify this for all appropriate primes less than 100," so that's all you have to do. You are being told that it's true for every prime of the form $3k+1$, but you aren't being asked to prove that (which would be considerably harder to do).
$endgroup$
– Gerry Myerson
Jan 25 at 3:44




$begingroup$
The question, the way you have quoted it, says "verify this for all appropriate primes less than 100," so that's all you have to do. You are being told that it's true for every prime of the form $3k+1$, but you aren't being asked to prove that (which would be considerably harder to do).
$endgroup$
– Gerry Myerson
Jan 25 at 3:44




2




2




$begingroup$
Don't. It takes time to learn how to interpret mathematical questions. They're in a foreign language.
$endgroup$
– Gerry Myerson
Jan 25 at 3:59




$begingroup$
Don't. It takes time to learn how to interpret mathematical questions. They're in a foreign language.
$endgroup$
– Gerry Myerson
Jan 25 at 3:59










1 Answer
1






active

oldest

votes


















1












$begingroup$

Proof for all $p$'s.



Recall Thue's result.



Thue's Lemma: Let $n>1$ be an integer, and $a$ be an integer coprime to $n$. Then, there exists $x,y$ with $0<x,y<sqrt{n}$ so that, $xequiv aypmod{n}$.



In some sense, $(a,n)=1$ can be represented as a fraction, both of whose numerator and denominator is at most $sqrt{n}$. The proof is available in many places, and I omit herein.




Now, we will apply Thue's lemma. The key point is the well-known fact, if $pequiv 1pmod{6}$ a prime, then $-3$ is a quadratic residue modulo $p$ (that is, there exists an $x$ so that $x^2equiv -3pmod{p}$). Now, let $0<x'<p$ be such a number, with $x'<p/2$ (note that, both $x'$ and $p-x'$ are so, and one of them is less than $p/2$, which, I assume to be $x'$ without any loss of generality).



Now, there is $0<a,b<sqrt{p}$ with $aequiv x'bpmod{p}$. With this, $a^2+3b^2 equiv b^2(x'^2+3)equiv 0pmod{p}$. It now suffices to show $a^2+3b^2=p$. Clearly, $a^2+3b^2<4p$, hence the possibilities are $a^2+3b^2=p,2p,3p$. If $a^2+3b^2=3p$, then in modulo $3$ we see that $3mid a$. Letting $a=3a_1$, we have $3a_1^2+b^2=p$, and therefore, $p$ can be represented in the desired manner.



If $a^2+3b^2=2p$, then we observe that $a$ and $b$ are of same parity. In both cases, you may see that $a^2+3b^2equiv 0pmod{4}$, while $4nmid 2p$, a contradiction.



We are done.






share|cite|improve this answer









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    1












    $begingroup$

    Proof for all $p$'s.



    Recall Thue's result.



    Thue's Lemma: Let $n>1$ be an integer, and $a$ be an integer coprime to $n$. Then, there exists $x,y$ with $0<x,y<sqrt{n}$ so that, $xequiv aypmod{n}$.



    In some sense, $(a,n)=1$ can be represented as a fraction, both of whose numerator and denominator is at most $sqrt{n}$. The proof is available in many places, and I omit herein.




    Now, we will apply Thue's lemma. The key point is the well-known fact, if $pequiv 1pmod{6}$ a prime, then $-3$ is a quadratic residue modulo $p$ (that is, there exists an $x$ so that $x^2equiv -3pmod{p}$). Now, let $0<x'<p$ be such a number, with $x'<p/2$ (note that, both $x'$ and $p-x'$ are so, and one of them is less than $p/2$, which, I assume to be $x'$ without any loss of generality).



    Now, there is $0<a,b<sqrt{p}$ with $aequiv x'bpmod{p}$. With this, $a^2+3b^2 equiv b^2(x'^2+3)equiv 0pmod{p}$. It now suffices to show $a^2+3b^2=p$. Clearly, $a^2+3b^2<4p$, hence the possibilities are $a^2+3b^2=p,2p,3p$. If $a^2+3b^2=3p$, then in modulo $3$ we see that $3mid a$. Letting $a=3a_1$, we have $3a_1^2+b^2=p$, and therefore, $p$ can be represented in the desired manner.



    If $a^2+3b^2=2p$, then we observe that $a$ and $b$ are of same parity. In both cases, you may see that $a^2+3b^2equiv 0pmod{4}$, while $4nmid 2p$, a contradiction.



    We are done.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Proof for all $p$'s.



      Recall Thue's result.



      Thue's Lemma: Let $n>1$ be an integer, and $a$ be an integer coprime to $n$. Then, there exists $x,y$ with $0<x,y<sqrt{n}$ so that, $xequiv aypmod{n}$.



      In some sense, $(a,n)=1$ can be represented as a fraction, both of whose numerator and denominator is at most $sqrt{n}$. The proof is available in many places, and I omit herein.




      Now, we will apply Thue's lemma. The key point is the well-known fact, if $pequiv 1pmod{6}$ a prime, then $-3$ is a quadratic residue modulo $p$ (that is, there exists an $x$ so that $x^2equiv -3pmod{p}$). Now, let $0<x'<p$ be such a number, with $x'<p/2$ (note that, both $x'$ and $p-x'$ are so, and one of them is less than $p/2$, which, I assume to be $x'$ without any loss of generality).



      Now, there is $0<a,b<sqrt{p}$ with $aequiv x'bpmod{p}$. With this, $a^2+3b^2 equiv b^2(x'^2+3)equiv 0pmod{p}$. It now suffices to show $a^2+3b^2=p$. Clearly, $a^2+3b^2<4p$, hence the possibilities are $a^2+3b^2=p,2p,3p$. If $a^2+3b^2=3p$, then in modulo $3$ we see that $3mid a$. Letting $a=3a_1$, we have $3a_1^2+b^2=p$, and therefore, $p$ can be represented in the desired manner.



      If $a^2+3b^2=2p$, then we observe that $a$ and $b$ are of same parity. In both cases, you may see that $a^2+3b^2equiv 0pmod{4}$, while $4nmid 2p$, a contradiction.



      We are done.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Proof for all $p$'s.



        Recall Thue's result.



        Thue's Lemma: Let $n>1$ be an integer, and $a$ be an integer coprime to $n$. Then, there exists $x,y$ with $0<x,y<sqrt{n}$ so that, $xequiv aypmod{n}$.



        In some sense, $(a,n)=1$ can be represented as a fraction, both of whose numerator and denominator is at most $sqrt{n}$. The proof is available in many places, and I omit herein.




        Now, we will apply Thue's lemma. The key point is the well-known fact, if $pequiv 1pmod{6}$ a prime, then $-3$ is a quadratic residue modulo $p$ (that is, there exists an $x$ so that $x^2equiv -3pmod{p}$). Now, let $0<x'<p$ be such a number, with $x'<p/2$ (note that, both $x'$ and $p-x'$ are so, and one of them is less than $p/2$, which, I assume to be $x'$ without any loss of generality).



        Now, there is $0<a,b<sqrt{p}$ with $aequiv x'bpmod{p}$. With this, $a^2+3b^2 equiv b^2(x'^2+3)equiv 0pmod{p}$. It now suffices to show $a^2+3b^2=p$. Clearly, $a^2+3b^2<4p$, hence the possibilities are $a^2+3b^2=p,2p,3p$. If $a^2+3b^2=3p$, then in modulo $3$ we see that $3mid a$. Letting $a=3a_1$, we have $3a_1^2+b^2=p$, and therefore, $p$ can be represented in the desired manner.



        If $a^2+3b^2=2p$, then we observe that $a$ and $b$ are of same parity. In both cases, you may see that $a^2+3b^2equiv 0pmod{4}$, while $4nmid 2p$, a contradiction.



        We are done.






        share|cite|improve this answer









        $endgroup$



        Proof for all $p$'s.



        Recall Thue's result.



        Thue's Lemma: Let $n>1$ be an integer, and $a$ be an integer coprime to $n$. Then, there exists $x,y$ with $0<x,y<sqrt{n}$ so that, $xequiv aypmod{n}$.



        In some sense, $(a,n)=1$ can be represented as a fraction, both of whose numerator and denominator is at most $sqrt{n}$. The proof is available in many places, and I omit herein.




        Now, we will apply Thue's lemma. The key point is the well-known fact, if $pequiv 1pmod{6}$ a prime, then $-3$ is a quadratic residue modulo $p$ (that is, there exists an $x$ so that $x^2equiv -3pmod{p}$). Now, let $0<x'<p$ be such a number, with $x'<p/2$ (note that, both $x'$ and $p-x'$ are so, and one of them is less than $p/2$, which, I assume to be $x'$ without any loss of generality).



        Now, there is $0<a,b<sqrt{p}$ with $aequiv x'bpmod{p}$. With this, $a^2+3b^2 equiv b^2(x'^2+3)equiv 0pmod{p}$. It now suffices to show $a^2+3b^2=p$. Clearly, $a^2+3b^2<4p$, hence the possibilities are $a^2+3b^2=p,2p,3p$. If $a^2+3b^2=3p$, then in modulo $3$ we see that $3mid a$. Letting $a=3a_1$, we have $3a_1^2+b^2=p$, and therefore, $p$ can be represented in the desired manner.



        If $a^2+3b^2=2p$, then we observe that $a$ and $b$ are of same parity. In both cases, you may see that $a^2+3b^2equiv 0pmod{4}$, while $4nmid 2p$, a contradiction.



        We are done.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 25 at 8:07









        AaronAaron

        1,912415




        1,912415






























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