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

Question

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!

Answer 1

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:

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

Answer 2

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

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