Is it true that $(a^2-ab+b^2)(c^2-cd+d^2)=h^2-hk+k^2$ for some coprime $h$ and $k$?












3












$begingroup$



Let us consider two numbers of the form $a^2 - ab + b^2$ and $c^2 - cd + d^2$ which are not both divisible by $3$ and such that $(a, b) = 1$ and $(c,d) = 1$. Running some computations it seems that the product $$(a^2 -ab + b^2)(c^2 - cd + d^2) $$ is still of the form $h^2 - hk + k^2$ for some suitable coprime integers $h,k$. Is this true?




I tried to prove it by writing down explicitly the product and looking for patterns, but I had no luck. Any help would be appreciated!






elementary-number-theory complex-numbers quadratics







share|cite|improve this question









New contributor




Al Tac is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    The prime divisors of such numbers are of the form $6n+1$, see here. So the product $(6n+1)(6m+1)$ is again of this form.
    $endgroup$
    – Dietrich Burde
    7 hours ago












  • $begingroup$
    Yes but the product is not a prime anymore: are you saying that any number of the form $6x+1$ can be written as $h^2 - hk + k^2$? In the topic that you linked only the case of prime numbers is discussed, if I am not mistaken.
    $endgroup$
    – Al Tac
    6 hours ago








  • 1




    $begingroup$
    Yes, it is not a prime anymore, but a product of primes of the form $6n+1$. It is represented by the quadratic form $x^2-xy+y^2$.
    $endgroup$
    – Dietrich Burde
    6 hours ago


















3












$begingroup$



Let us consider two numbers of the form $a^2 - ab + b^2$ and $c^2 - cd + d^2$ which are not both divisible by $3$ and such that $(a, b) = 1$ and $(c,d) = 1$. Running some computations it seems that the product $$(a^2 -ab + b^2)(c^2 - cd + d^2) $$ is still of the form $h^2 - hk + k^2$ for some suitable coprime integers $h,k$. Is this true?




I tried to prove it by writing down explicitly the product and looking for patterns, but I had no luck. Any help would be appreciated!






elementary-number-theory complex-numbers quadratics







share|cite|improve this question









New contributor




Al Tac is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    The prime divisors of such numbers are of the form $6n+1$, see here. So the product $(6n+1)(6m+1)$ is again of this form.
    $endgroup$
    – Dietrich Burde
    7 hours ago












  • $begingroup$
    Yes but the product is not a prime anymore: are you saying that any number of the form $6x+1$ can be written as $h^2 - hk + k^2$? In the topic that you linked only the case of prime numbers is discussed, if I am not mistaken.
    $endgroup$
    – Al Tac
    6 hours ago








  • 1




    $begingroup$
    Yes, it is not a prime anymore, but a product of primes of the form $6n+1$. It is represented by the quadratic form $x^2-xy+y^2$.
    $endgroup$
    – Dietrich Burde
    6 hours ago
















3












3








3


0



$begingroup$



Let us consider two numbers of the form $a^2 - ab + b^2$ and $c^2 - cd + d^2$ which are not both divisible by $3$ and such that $(a, b) = 1$ and $(c,d) = 1$. Running some computations it seems that the product $$(a^2 -ab + b^2)(c^2 - cd + d^2) $$ is still of the form $h^2 - hk + k^2$ for some suitable coprime integers $h,k$. Is this true?




I tried to prove it by writing down explicitly the product and looking for patterns, but I had no luck. Any help would be appreciated!






elementary-number-theory complex-numbers quadratics







share|cite|improve this question









New contributor




Al Tac is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$





Let us consider two numbers of the form $a^2 - ab + b^2$ and $c^2 - cd + d^2$ which are not both divisible by $3$ and such that $(a, b) = 1$ and $(c,d) = 1$. Running some computations it seems that the product $$(a^2 -ab + b^2)(c^2 - cd + d^2) $$ is still of the form $h^2 - hk + k^2$ for some suitable coprime integers $h,k$. Is this true?




I tried to prove it by writing down explicitly the product and looking for patterns, but I had no luck. Any help would be appreciated!






elementary-number-theory complex-numbers quadratics




elementary-number-theory complex-numbers quadratics






share|cite|improve this question









New contributor




Al Tac is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Al Tac is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









Asaf Karagila♦

306k33438769




306k33438769






New contributor




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Check out our Code of Conduct.









asked 8 hours ago









Al TacAl Tac

192




192




New contributor




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New contributor





Al Tac is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Al Tac is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    The prime divisors of such numbers are of the form $6n+1$, see here. So the product $(6n+1)(6m+1)$ is again of this form.
    $endgroup$
    – Dietrich Burde
    7 hours ago












  • $begingroup$
    Yes but the product is not a prime anymore: are you saying that any number of the form $6x+1$ can be written as $h^2 - hk + k^2$? In the topic that you linked only the case of prime numbers is discussed, if I am not mistaken.
    $endgroup$
    – Al Tac
    6 hours ago








  • 1




    $begingroup$
    Yes, it is not a prime anymore, but a product of primes of the form $6n+1$. It is represented by the quadratic form $x^2-xy+y^2$.
    $endgroup$
    – Dietrich Burde
    6 hours ago




















  • $begingroup$
    The prime divisors of such numbers are of the form $6n+1$, see here. So the product $(6n+1)(6m+1)$ is again of this form.
    $endgroup$
    – Dietrich Burde
    7 hours ago












  • $begingroup$
    Yes but the product is not a prime anymore: are you saying that any number of the form $6x+1$ can be written as $h^2 - hk + k^2$? In the topic that you linked only the case of prime numbers is discussed, if I am not mistaken.
    $endgroup$
    – Al Tac
    6 hours ago








  • 1




    $begingroup$
    Yes, it is not a prime anymore, but a product of primes of the form $6n+1$. It is represented by the quadratic form $x^2-xy+y^2$.
    $endgroup$
    – Dietrich Burde
    6 hours ago


















$begingroup$
The prime divisors of such numbers are of the form $6n+1$, see here. So the product $(6n+1)(6m+1)$ is again of this form.
$endgroup$
– Dietrich Burde
7 hours ago






$begingroup$
The prime divisors of such numbers are of the form $6n+1$, see here. So the product $(6n+1)(6m+1)$ is again of this form.
$endgroup$
– Dietrich Burde
7 hours ago














$begingroup$
Yes but the product is not a prime anymore: are you saying that any number of the form $6x+1$ can be written as $h^2 - hk + k^2$? In the topic that you linked only the case of prime numbers is discussed, if I am not mistaken.
$endgroup$
– Al Tac
6 hours ago






$begingroup$
Yes but the product is not a prime anymore: are you saying that any number of the form $6x+1$ can be written as $h^2 - hk + k^2$? In the topic that you linked only the case of prime numbers is discussed, if I am not mistaken.
$endgroup$
– Al Tac
6 hours ago






1




1




$begingroup$
Yes, it is not a prime anymore, but a product of primes of the form $6n+1$. It is represented by the quadratic form $x^2-xy+y^2$.
$endgroup$
– Dietrich Burde
6 hours ago






$begingroup$
Yes, it is not a prime anymore, but a product of primes of the form $6n+1$. It is represented by the quadratic form $x^2-xy+y^2$.
$endgroup$
– Dietrich Burde
6 hours ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

Let $u$ and $v$ be zeros of $x^2-x+1=0$. Then $$x^2-x+1 = (x-u)(x-v)$$ and $$u^2=u-1;;;wedge ;;;;v^2=v-1$$



so $$(a^2 -ab + b^2)(c^2 - cd + d^2) = color{red}{(a-bu)}color{blue}{(a-bv)}color{red}{(c- du)}color{blue}{(c-dv)}$$
$$= color{red}{Big(ac+bdu^2-(ad+bc)uBig)}color{blue}{Big(ac+bdv^2-(ad+bc)vBig)}$$
$$= Big(underbrace{ac-bd}_m-underbrace{(ad+bc-bd)}_n uBig)Big(underbrace{ac-bd}_m-underbrace{(ad+bc-bd)}_n vBig)$$



$$ =(m-nu)(m-nv) = m^2-mn+n^2$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Very nice how you bring in $u,vin mathbb C$ for manipulation purposes, and they drop right out at the end.
    $endgroup$
    – Keith Backman
    5 hours ago










  • $begingroup$
    Very nice indeed, thank you! Is it obvious that $m$ and $n$ are still coprime? Also, what happens if both $a^2 - ab + b^2$ and $c^2 - cd + d^2$ are divisible by $3$? Something should go wrong, because numbers of the form $m^2 - mn + n^2$ are never divisible by $9$ if $m$ and $n$ are coprime, right?
    $endgroup$
    – Al Tac
    4 hours ago










  • $begingroup$
    I'm sorry I'm not sure if they are relatively prime.
    $endgroup$
    – Maria Mazur
    4 hours ago



















2












$begingroup$

There is this Identity:



$[(ac+bd)^2-(ab(c^2+d^2)-(abcd)+cd(a^2+b^2))+(bc-ad)^2]=(a^2-ab+b^2)(c^2-cd-d^2)$



Hence for:



$(a^2-ab+b^2)(c^2-cd-d^2)=(h^2-hk+k^2)$



$h=(ac+bd)$



$k=(bc-ad)$



$hk=(ac+bd)(bc-ad)$



Condition (c,d)=(2b,b-2a)



For $(a,b,c,d)=(3,7,14,1)$ we get:



$(49^2-49*95+95^2)=(37)*(183)=6771$






share|cite|improve this answer










New contributor




Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    I am sorry, I am not sure I follow. What is your definition of $k$? If I take $h = ac + bd$ and $k = bc - ad$, in $h^2 - hk + k^2$ there is no term $abcd$, right? Also, in your example $k$ would be $-5$ not $+5$, so you would obtain $14^2 + 130 + 5^2$, it seems.
    $endgroup$
    – Al Tac
    6 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Let $u$ and $v$ be zeros of $x^2-x+1=0$. Then $$x^2-x+1 = (x-u)(x-v)$$ and $$u^2=u-1;;;wedge ;;;;v^2=v-1$$



so $$(a^2 -ab + b^2)(c^2 - cd + d^2) = color{red}{(a-bu)}color{blue}{(a-bv)}color{red}{(c- du)}color{blue}{(c-dv)}$$
$$= color{red}{Big(ac+bdu^2-(ad+bc)uBig)}color{blue}{Big(ac+bdv^2-(ad+bc)vBig)}$$
$$= Big(underbrace{ac-bd}_m-underbrace{(ad+bc-bd)}_n uBig)Big(underbrace{ac-bd}_m-underbrace{(ad+bc-bd)}_n vBig)$$



$$ =(m-nu)(m-nv) = m^2-mn+n^2$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Very nice how you bring in $u,vin mathbb C$ for manipulation purposes, and they drop right out at the end.
    $endgroup$
    – Keith Backman
    5 hours ago










  • $begingroup$
    Very nice indeed, thank you! Is it obvious that $m$ and $n$ are still coprime? Also, what happens if both $a^2 - ab + b^2$ and $c^2 - cd + d^2$ are divisible by $3$? Something should go wrong, because numbers of the form $m^2 - mn + n^2$ are never divisible by $9$ if $m$ and $n$ are coprime, right?
    $endgroup$
    – Al Tac
    4 hours ago










  • $begingroup$
    I'm sorry I'm not sure if they are relatively prime.
    $endgroup$
    – Maria Mazur
    4 hours ago
















3












$begingroup$

Let $u$ and $v$ be zeros of $x^2-x+1=0$. Then $$x^2-x+1 = (x-u)(x-v)$$ and $$u^2=u-1;;;wedge ;;;;v^2=v-1$$



so $$(a^2 -ab + b^2)(c^2 - cd + d^2) = color{red}{(a-bu)}color{blue}{(a-bv)}color{red}{(c- du)}color{blue}{(c-dv)}$$
$$= color{red}{Big(ac+bdu^2-(ad+bc)uBig)}color{blue}{Big(ac+bdv^2-(ad+bc)vBig)}$$
$$= Big(underbrace{ac-bd}_m-underbrace{(ad+bc-bd)}_n uBig)Big(underbrace{ac-bd}_m-underbrace{(ad+bc-bd)}_n vBig)$$



$$ =(m-nu)(m-nv) = m^2-mn+n^2$$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Very nice how you bring in $u,vin mathbb C$ for manipulation purposes, and they drop right out at the end.
    $endgroup$
    – Keith Backman
    5 hours ago










  • $begingroup$
    Very nice indeed, thank you! Is it obvious that $m$ and $n$ are still coprime? Also, what happens if both $a^2 - ab + b^2$ and $c^2 - cd + d^2$ are divisible by $3$? Something should go wrong, because numbers of the form $m^2 - mn + n^2$ are never divisible by $9$ if $m$ and $n$ are coprime, right?
    $endgroup$
    – Al Tac
    4 hours ago










  • $begingroup$
    I'm sorry I'm not sure if they are relatively prime.
    $endgroup$
    – Maria Mazur
    4 hours ago














3












3








3





$begingroup$

Let $u$ and $v$ be zeros of $x^2-x+1=0$. Then $$x^2-x+1 = (x-u)(x-v)$$ and $$u^2=u-1;;;wedge ;;;;v^2=v-1$$



so $$(a^2 -ab + b^2)(c^2 - cd + d^2) = color{red}{(a-bu)}color{blue}{(a-bv)}color{red}{(c- du)}color{blue}{(c-dv)}$$
$$= color{red}{Big(ac+bdu^2-(ad+bc)uBig)}color{blue}{Big(ac+bdv^2-(ad+bc)vBig)}$$
$$= Big(underbrace{ac-bd}_m-underbrace{(ad+bc-bd)}_n uBig)Big(underbrace{ac-bd}_m-underbrace{(ad+bc-bd)}_n vBig)$$



$$ =(m-nu)(m-nv) = m^2-mn+n^2$$






share|cite|improve this answer









$endgroup$



Let $u$ and $v$ be zeros of $x^2-x+1=0$. Then $$x^2-x+1 = (x-u)(x-v)$$ and $$u^2=u-1;;;wedge ;;;;v^2=v-1$$



so $$(a^2 -ab + b^2)(c^2 - cd + d^2) = color{red}{(a-bu)}color{blue}{(a-bv)}color{red}{(c- du)}color{blue}{(c-dv)}$$
$$= color{red}{Big(ac+bdu^2-(ad+bc)uBig)}color{blue}{Big(ac+bdv^2-(ad+bc)vBig)}$$
$$= Big(underbrace{ac-bd}_m-underbrace{(ad+bc-bd)}_n uBig)Big(underbrace{ac-bd}_m-underbrace{(ad+bc-bd)}_n vBig)$$



$$ =(m-nu)(m-nv) = m^2-mn+n^2$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 6 hours ago









Maria MazurMaria Mazur

46.9k1260120




46.9k1260120








  • 1




    $begingroup$
    Very nice how you bring in $u,vin mathbb C$ for manipulation purposes, and they drop right out at the end.
    $endgroup$
    – Keith Backman
    5 hours ago










  • $begingroup$
    Very nice indeed, thank you! Is it obvious that $m$ and $n$ are still coprime? Also, what happens if both $a^2 - ab + b^2$ and $c^2 - cd + d^2$ are divisible by $3$? Something should go wrong, because numbers of the form $m^2 - mn + n^2$ are never divisible by $9$ if $m$ and $n$ are coprime, right?
    $endgroup$
    – Al Tac
    4 hours ago










  • $begingroup$
    I'm sorry I'm not sure if they are relatively prime.
    $endgroup$
    – Maria Mazur
    4 hours ago














  • 1




    $begingroup$
    Very nice how you bring in $u,vin mathbb C$ for manipulation purposes, and they drop right out at the end.
    $endgroup$
    – Keith Backman
    5 hours ago










  • $begingroup$
    Very nice indeed, thank you! Is it obvious that $m$ and $n$ are still coprime? Also, what happens if both $a^2 - ab + b^2$ and $c^2 - cd + d^2$ are divisible by $3$? Something should go wrong, because numbers of the form $m^2 - mn + n^2$ are never divisible by $9$ if $m$ and $n$ are coprime, right?
    $endgroup$
    – Al Tac
    4 hours ago










  • $begingroup$
    I'm sorry I'm not sure if they are relatively prime.
    $endgroup$
    – Maria Mazur
    4 hours ago








1




1




$begingroup$
Very nice how you bring in $u,vin mathbb C$ for manipulation purposes, and they drop right out at the end.
$endgroup$
– Keith Backman
5 hours ago




$begingroup$
Very nice how you bring in $u,vin mathbb C$ for manipulation purposes, and they drop right out at the end.
$endgroup$
– Keith Backman
5 hours ago












$begingroup$
Very nice indeed, thank you! Is it obvious that $m$ and $n$ are still coprime? Also, what happens if both $a^2 - ab + b^2$ and $c^2 - cd + d^2$ are divisible by $3$? Something should go wrong, because numbers of the form $m^2 - mn + n^2$ are never divisible by $9$ if $m$ and $n$ are coprime, right?
$endgroup$
– Al Tac
4 hours ago




$begingroup$
Very nice indeed, thank you! Is it obvious that $m$ and $n$ are still coprime? Also, what happens if both $a^2 - ab + b^2$ and $c^2 - cd + d^2$ are divisible by $3$? Something should go wrong, because numbers of the form $m^2 - mn + n^2$ are never divisible by $9$ if $m$ and $n$ are coprime, right?
$endgroup$
– Al Tac
4 hours ago












$begingroup$
I'm sorry I'm not sure if they are relatively prime.
$endgroup$
– Maria Mazur
4 hours ago




$begingroup$
I'm sorry I'm not sure if they are relatively prime.
$endgroup$
– Maria Mazur
4 hours ago











2












$begingroup$

There is this Identity:



$[(ac+bd)^2-(ab(c^2+d^2)-(abcd)+cd(a^2+b^2))+(bc-ad)^2]=(a^2-ab+b^2)(c^2-cd-d^2)$



Hence for:



$(a^2-ab+b^2)(c^2-cd-d^2)=(h^2-hk+k^2)$



$h=(ac+bd)$



$k=(bc-ad)$



$hk=(ac+bd)(bc-ad)$



Condition (c,d)=(2b,b-2a)



For $(a,b,c,d)=(3,7,14,1)$ we get:



$(49^2-49*95+95^2)=(37)*(183)=6771$






share|cite|improve this answer










New contributor




Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    I am sorry, I am not sure I follow. What is your definition of $k$? If I take $h = ac + bd$ and $k = bc - ad$, in $h^2 - hk + k^2$ there is no term $abcd$, right? Also, in your example $k$ would be $-5$ not $+5$, so you would obtain $14^2 + 130 + 5^2$, it seems.
    $endgroup$
    – Al Tac
    6 hours ago
















2












$begingroup$

There is this Identity:



$[(ac+bd)^2-(ab(c^2+d^2)-(abcd)+cd(a^2+b^2))+(bc-ad)^2]=(a^2-ab+b^2)(c^2-cd-d^2)$



Hence for:



$(a^2-ab+b^2)(c^2-cd-d^2)=(h^2-hk+k^2)$



$h=(ac+bd)$



$k=(bc-ad)$



$hk=(ac+bd)(bc-ad)$



Condition (c,d)=(2b,b-2a)



For $(a,b,c,d)=(3,7,14,1)$ we get:



$(49^2-49*95+95^2)=(37)*(183)=6771$






share|cite|improve this answer










New contributor




Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$













  • $begingroup$
    I am sorry, I am not sure I follow. What is your definition of $k$? If I take $h = ac + bd$ and $k = bc - ad$, in $h^2 - hk + k^2$ there is no term $abcd$, right? Also, in your example $k$ would be $-5$ not $+5$, so you would obtain $14^2 + 130 + 5^2$, it seems.
    $endgroup$
    – Al Tac
    6 hours ago














2












2








2





$begingroup$

There is this Identity:



$[(ac+bd)^2-(ab(c^2+d^2)-(abcd)+cd(a^2+b^2))+(bc-ad)^2]=(a^2-ab+b^2)(c^2-cd-d^2)$



Hence for:



$(a^2-ab+b^2)(c^2-cd-d^2)=(h^2-hk+k^2)$



$h=(ac+bd)$



$k=(bc-ad)$



$hk=(ac+bd)(bc-ad)$



Condition (c,d)=(2b,b-2a)



For $(a,b,c,d)=(3,7,14,1)$ we get:



$(49^2-49*95+95^2)=(37)*(183)=6771$






share|cite|improve this answer










New contributor




Sam is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$



There is this Identity:



$[(ac+bd)^2-(ab(c^2+d^2)-(abcd)+cd(a^2+b^2))+(bc-ad)^2]=(a^2-ab+b^2)(c^2-cd-d^2)$



Hence for:



$(a^2-ab+b^2)(c^2-cd-d^2)=(h^2-hk+k^2)$



$h=(ac+bd)$



$k=(bc-ad)$



$hk=(ac+bd)(bc-ad)$



Condition (c,d)=(2b,b-2a)



For $(a,b,c,d)=(3,7,14,1)$ we get:



$(49^2-49*95+95^2)=(37)*(183)=6771$







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edited 6 hours ago





















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  • $begingroup$
    I am sorry, I am not sure I follow. What is your definition of $k$? If I take $h = ac + bd$ and $k = bc - ad$, in $h^2 - hk + k^2$ there is no term $abcd$, right? Also, in your example $k$ would be $-5$ not $+5$, so you would obtain $14^2 + 130 + 5^2$, it seems.
    $endgroup$
    – Al Tac
    6 hours ago


















  • $begingroup$
    I am sorry, I am not sure I follow. What is your definition of $k$? If I take $h = ac + bd$ and $k = bc - ad$, in $h^2 - hk + k^2$ there is no term $abcd$, right? Also, in your example $k$ would be $-5$ not $+5$, so you would obtain $14^2 + 130 + 5^2$, it seems.
    $endgroup$
    – Al Tac
    6 hours ago
















$begingroup$
I am sorry, I am not sure I follow. What is your definition of $k$? If I take $h = ac + bd$ and $k = bc - ad$, in $h^2 - hk + k^2$ there is no term $abcd$, right? Also, in your example $k$ would be $-5$ not $+5$, so you would obtain $14^2 + 130 + 5^2$, it seems.
$endgroup$
– Al Tac
6 hours ago




$begingroup$
I am sorry, I am not sure I follow. What is your definition of $k$? If I take $h = ac + bd$ and $k = bc - ad$, in $h^2 - hk + k^2$ there is no term $abcd$, right? Also, in your example $k$ would be $-5$ not $+5$, so you would obtain $14^2 + 130 + 5^2$, it seems.
$endgroup$
– Al Tac
6 hours ago










Al Tac is a new contributor. Be nice, and check out our Code of Conduct.










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