Is it true that $(a^2-ab+b^2)(c^2-cd+d^2)=h^2-hk+k^2$ for some coprime $h$ and $k$?$x^2-xy+y^2=n$ how many solutionsFor $a, b$ coprime, if $p geq 5$ is a prime which divides $a^2 - ab + b^2$, then $p equiv 1 pmod6$Showing $aperp b$ and $nne 0$ implies $a+bkperp n$ for some $k$If $x=123456789101112131415161718$, then $xequiv 6pmod16$ and $xequiv 0pmod 6$Are there long arithmetic progressions non-coprime with the given number?Solution to some confusing complex equationProof that the product of primitive Pythagorean hypotenuses is also a primitive Pythagorean hypotenuseIf all pairs of addends that sum up to $N$ are coprime, then $N$ is prime.Prove that $a$ and $b$ are coprime whenever $a+b$ and $a-b$ are coprimeMaximal Consecutive Integer SequenceIs the sum of two coprime natural numbers prime?Probability of k random integers being coprimes

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


$x^2-xy+y^2=n$ how many solutionsFor $a, b$ coprime, if $p geq 5$ is a prime which divides $a^2 - ab + b^2$, then $p equiv 1 pmod6$Showing $aperp b$ and $nne 0$ implies $a+bkperp n$ for some $k$If $x=123456789101112131415161718$, then $xequiv 6pmod16$ and $xequiv 0pmod 6$Are there long arithmetic progressions non-coprime with the given number?Solution to some confusing complex equationProof that the product of primitive Pythagorean hypotenuses is also a primitive Pythagorean hypotenuseIf all pairs of addends that sum up to $N$ are coprime, then $N$ is prime.Prove that $a$ and $b$ are coprime whenever $a+b$ and $a-b$ are coprimeMaximal Consecutive Integer SequenceIs the sum of two coprime natural numbers prime?Probability of k random integers being coprimes













4












$begingroup$



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!










share|cite|improve this question











$endgroup$











  • $begingroup$
    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
    Mar 14 at 16:20











  • $begingroup$
    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
    Mar 14 at 17:33







  • 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
    Mar 14 at 17:56











  • $begingroup$
    Thanks! I found a complete answer to the question of which numbers $n$ are represented as $h^2 - hk + k^2$ with $(h,k) = 1$ in Problem 3 of Section 3.7 of Niven-Zuckerman-Montgomery, An introduction to the theory of numbers. They are indeed the ones of the form $n = 3^alpha prod_i p_i^e_i$ with $alpha = 0, 1$ and $p_i equiv 1 pmod6$ for all $i$.
    $endgroup$
    – Al Tac
    Mar 15 at 8:24















4












$begingroup$



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!










share|cite|improve this question











$endgroup$











  • $begingroup$
    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
    Mar 14 at 16:20











  • $begingroup$
    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
    Mar 14 at 17:33







  • 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
    Mar 14 at 17:56











  • $begingroup$
    Thanks! I found a complete answer to the question of which numbers $n$ are represented as $h^2 - hk + k^2$ with $(h,k) = 1$ in Problem 3 of Section 3.7 of Niven-Zuckerman-Montgomery, An introduction to the theory of numbers. They are indeed the ones of the form $n = 3^alpha prod_i p_i^e_i$ with $alpha = 0, 1$ and $p_i equiv 1 pmod6$ for all $i$.
    $endgroup$
    – Al Tac
    Mar 15 at 8:24













4












4








4


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!










share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 14 at 20:30









Asaf Karagila

306k33438769




306k33438769










asked Mar 14 at 15:35









Al TacAl Tac

242




242











  • $begingroup$
    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
    Mar 14 at 16:20











  • $begingroup$
    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
    Mar 14 at 17:33







  • 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
    Mar 14 at 17:56











  • $begingroup$
    Thanks! I found a complete answer to the question of which numbers $n$ are represented as $h^2 - hk + k^2$ with $(h,k) = 1$ in Problem 3 of Section 3.7 of Niven-Zuckerman-Montgomery, An introduction to the theory of numbers. They are indeed the ones of the form $n = 3^alpha prod_i p_i^e_i$ with $alpha = 0, 1$ and $p_i equiv 1 pmod6$ for all $i$.
    $endgroup$
    – Al Tac
    Mar 15 at 8:24
















  • $begingroup$
    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
    Mar 14 at 16:20











  • $begingroup$
    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
    Mar 14 at 17:33







  • 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
    Mar 14 at 17:56











  • $begingroup$
    Thanks! I found a complete answer to the question of which numbers $n$ are represented as $h^2 - hk + k^2$ with $(h,k) = 1$ in Problem 3 of Section 3.7 of Niven-Zuckerman-Montgomery, An introduction to the theory of numbers. They are indeed the ones of the form $n = 3^alpha prod_i p_i^e_i$ with $alpha = 0, 1$ and $p_i equiv 1 pmod6$ for all $i$.
    $endgroup$
    – Al Tac
    Mar 15 at 8:24















$begingroup$
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
Mar 14 at 16:20





$begingroup$
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
Mar 14 at 16:20













$begingroup$
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
Mar 14 at 17:33





$begingroup$
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
Mar 14 at 17:33





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
Mar 14 at 17:56





$begingroup$
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
Mar 14 at 17:56













$begingroup$
Thanks! I found a complete answer to the question of which numbers $n$ are represented as $h^2 - hk + k^2$ with $(h,k) = 1$ in Problem 3 of Section 3.7 of Niven-Zuckerman-Montgomery, An introduction to the theory of numbers. They are indeed the ones of the form $n = 3^alpha prod_i p_i^e_i$ with $alpha = 0, 1$ and $p_i equiv 1 pmod6$ for all $i$.
$endgroup$
– Al Tac
Mar 15 at 8:24




$begingroup$
Thanks! I found a complete answer to the question of which numbers $n$ are represented as $h^2 - hk + k^2$ with $(h,k) = 1$ in Problem 3 of Section 3.7 of Niven-Zuckerman-Montgomery, An introduction to the theory of numbers. They are indeed the ones of the form $n = 3^alpha prod_i p_i^e_i$ with $alpha = 0, 1$ and $p_i equiv 1 pmod6$ for all $i$.
$endgroup$
– Al Tac
Mar 15 at 8:24










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 $$(a^2 -ab + b^2)(c^2 - cd + d^2) = colorred(a-bu)colorblue(a-bv)colorred(c- du)colorblue(c-dv)$$
$$= colorredBig(ac+bdu^2-(ad+bc)uBig)colorblueBig(ac+bdv^2-(ad+bc)vBig)$$
$$= Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n uBig)Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n vBig)$$



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






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
    Mar 14 at 18:52










  • $begingroup$
    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
    Mar 14 at 19:25










  • $begingroup$
    I'm sorry I'm not sure if they are relatively prime.
    $endgroup$
    – Maria Mazur
    Mar 14 at 19:38


















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











$endgroup$












  • $begingroup$
    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
    Mar 14 at 17:41










  • $begingroup$
    @Al Tac. There was a typo. See the edited answer above.
    $endgroup$
    – Sam
    Mar 15 at 1:33










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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 $$(a^2 -ab + b^2)(c^2 - cd + d^2) = colorred(a-bu)colorblue(a-bv)colorred(c- du)colorblue(c-dv)$$
$$= colorredBig(ac+bdu^2-(ad+bc)uBig)colorblueBig(ac+bdv^2-(ad+bc)vBig)$$
$$= Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n uBig)Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n vBig)$$



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






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
    Mar 14 at 18:52










  • $begingroup$
    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
    Mar 14 at 19:25










  • $begingroup$
    I'm sorry I'm not sure if they are relatively prime.
    $endgroup$
    – Maria Mazur
    Mar 14 at 19:38















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 $$(a^2 -ab + b^2)(c^2 - cd + d^2) = colorred(a-bu)colorblue(a-bv)colorred(c- du)colorblue(c-dv)$$
$$= colorredBig(ac+bdu^2-(ad+bc)uBig)colorblueBig(ac+bdv^2-(ad+bc)vBig)$$
$$= Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n uBig)Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n vBig)$$



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






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
    Mar 14 at 18:52










  • $begingroup$
    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
    Mar 14 at 19:25










  • $begingroup$
    I'm sorry I'm not sure if they are relatively prime.
    $endgroup$
    – Maria Mazur
    Mar 14 at 19:38













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 $$(a^2 -ab + b^2)(c^2 - cd + d^2) = colorred(a-bu)colorblue(a-bv)colorred(c- du)colorblue(c-dv)$$
$$= colorredBig(ac+bdu^2-(ad+bc)uBig)colorblueBig(ac+bdv^2-(ad+bc)vBig)$$
$$= Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n uBig)Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n vBig)$$



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






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 $$(a^2 -ab + b^2)(c^2 - cd + d^2) = colorred(a-bu)colorblue(a-bv)colorred(c- du)colorblue(c-dv)$$
$$= colorredBig(ac+bdu^2-(ad+bc)uBig)colorblueBig(ac+bdv^2-(ad+bc)vBig)$$
$$= Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n uBig)Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n vBig)$$



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 14 at 18:01









Maria MazurMaria Mazur

47.9k1260120




47.9k1260120







  • 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
    Mar 14 at 18:52










  • $begingroup$
    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
    Mar 14 at 19:25










  • $begingroup$
    I'm sorry I'm not sure if they are relatively prime.
    $endgroup$
    – Maria Mazur
    Mar 14 at 19:38












  • 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
    Mar 14 at 18:52










  • $begingroup$
    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
    Mar 14 at 19:25










  • $begingroup$
    I'm sorry I'm not sure if they are relatively prime.
    $endgroup$
    – Maria Mazur
    Mar 14 at 19:38







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
Mar 14 at 18:52




$begingroup$
Very nice how you bring in $u,vin mathbb C$ for manipulation purposes, and they drop right out at the end.
$endgroup$
– Keith Backman
Mar 14 at 18:52












$begingroup$
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
Mar 14 at 19:25




$begingroup$
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
Mar 14 at 19:25












$begingroup$
I'm sorry I'm not sure if they are relatively prime.
$endgroup$
– Maria Mazur
Mar 14 at 19:38




$begingroup$
I'm sorry I'm not sure if they are relatively prime.
$endgroup$
– Maria Mazur
Mar 14 at 19:38











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











$endgroup$












  • $begingroup$
    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
    Mar 14 at 17:41










  • $begingroup$
    @Al Tac. There was a typo. See the edited answer above.
    $endgroup$
    – Sam
    Mar 15 at 1:33















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











$endgroup$












  • $begingroup$
    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
    Mar 14 at 17:41










  • $begingroup$
    @Al Tac. There was a typo. See the edited answer above.
    $endgroup$
    – Sam
    Mar 15 at 1:33













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











$endgroup$



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:



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







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 14 at 17:49

























answered Mar 14 at 16:55









SamSam

212




212











  • $begingroup$
    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
    Mar 14 at 17:41










  • $begingroup$
    @Al Tac. There was a typo. See the edited answer above.
    $endgroup$
    – Sam
    Mar 15 at 1:33
















  • $begingroup$
    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
    Mar 14 at 17:41










  • $begingroup$
    @Al Tac. There was a typo. See the edited answer above.
    $endgroup$
    – Sam
    Mar 15 at 1:33















$begingroup$
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
Mar 14 at 17:41




$begingroup$
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
Mar 14 at 17:41












$begingroup$
@Al Tac. There was a typo. See the edited answer above.
$endgroup$
– Sam
Mar 15 at 1:33




$begingroup$
@Al Tac. There was a typo. See the edited answer above.
$endgroup$
– Sam
Mar 15 at 1:33

















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