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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!

nonuser
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Al Tac
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  • 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. – Dietrich Burde Mar 14 '19 at 16:20
  • 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. – Al Tac Mar 14 '19 at 17:33
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    Yes, it is not a prime anymore, but a product of primes of the form $6n+1$. It is represented by the quadratic form $x^2-xy+y^2$. – Dietrich Burde Mar 14 '19 at 17:56
  • 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 $p_i \equiv 1 \pmod{6}$ for all $i$. – Al Tac Mar 15 '19 at 08:24

2 Answers2

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Let $u$ and $v$ be the roots 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 $$(a^2 -ab + b^2)(c^2 - cd + d^2) = \color{red}{(a-bu)}\color{blue}{(a-bv)}\color{red}{(c- du)}\color{blue}{(c-dv)}$$ $$= \color{red}{\Big(ac+bdu^2-(ad+bc)u\Big)}\color{blue}{\Big(ac+bdv^2-(ad+bc)v\Big)}$$ $$= \Big(\underbrace{ac-bd}_m-\underbrace{(ad+bc-bd)}_n u\Big)\Big(\underbrace{ac-bd}_m-\underbrace{(ad+bc-bd)}_n v\Big)$$

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

nonuser
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    Very nice how you bring in $u,v\in \mathbb C$ for manipulation purposes, and they drop right out at the end. – Keith Backman Mar 14 '19 at 18:52
  • 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? – Al Tac Mar 14 '19 at 19:25
  • I'm sorry I'm not sure if they are relatively prime. – nonuser Mar 14 '19 at 19:38
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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:

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

Sam
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  • 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. – Al Tac Mar 14 '19 at 17:41
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    @Al Tac. There was a typo. See the edited answer above. – Sam Mar 15 '19 at 01:33