Calculate $\gcd(a^2b^2, a^2 + ab + b^2)$

Question

Given $\gcd(a, b) = 1$, calculate $d =\gcd(a^2b^2, a^2 + ab + b^2)$ in terms of $a$ and $b$.

I have tried some manipulations of the terms arriving to some expressions such as that $d$ divides $a^4 + b^4$ or that $d$ divide s $(a+b)^4$ but those haven't given me much help. I have also tried dividing the problem in cases: $d$ divides $a$ but not $b$, $d$ divides $a$ and $b$, $d$ doesn't divide neither $a$ or $b$, etc

The first statements come from:

$d\vert a^2(a^2+ab+b^2) - (a^2b^2) = a^4 +a^3b$

$d\vert b^2(a^2+ab+b^2) - (a^2b^2) = ab^3 +b^4$

$d\vert ab(a^2+ab+b^2) - (a^2b^2) = a^3b+ab^3$

$d\vert a^4+a^3b - (a^3b+ab^3) + ab^3 +b^4= a^4 +b^4$

$d\vert a^4 +b^4= (a+b)^4-4(a^3b+ab^3)-6a^2b^2 \to d\vert (a+b)^4$

You might want to show some context behind your question. Where did your question come from? What did you try? These things can help other users provide you better insights that just a post with a statement of the problem and no more. — soupless, Mar 28 '22 at 08:30
What is $d$? Why $a^4+b^4$ or $(a+b)^4$? You should elaborate on your method by making edits to your post. — Kyan Cheung, Mar 28 '22 at 08:35
If there are additional information, please add it to the body by editing the post. — soupless, Mar 28 '22 at 08:35
Let $,c = a^j!+!abd!+!b^k.,$ Then $,(a,c) = (a,b^k) = 1,$ by $,(a,b)=1,$ & linked dupe. Similarly $(b,c) = 1,$ so $(a^mb^n,c) = 1$ again by dupe. — Bill Dubuque, Mar 28 '22 at 23:15

score 2 · Accepted Answer · answered Mar 28 '22 at 09:00

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It is easier in such problems to consider first the case when $d=p$ is a prime number (instead of an arbitary postive integer).

Indeed, suppose that $p\mid\gcd(a^2b^2,a^2+ab+b^2)$. Then, $p\mid a^2b^2$ and $p\mid a^2+ab+b^2$. Can you now find out for which prime $p$ it is possible given that $\gcd(a,b)=1$?

answered Mar 28 '22 at 09:00

richrow

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From this I can only assure that $d$ can't be prime, but how would I solve for an arbitrary natural number? – xImperiak Mar 28 '22 at 09:15
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Hence, there is no prime number $p$ that divides $\gcd(a^2b^2,a^2+ab+b^2)$, right? – richrow Mar 28 '22 at 09:45
Oooh, I see. Thank you – xImperiak Mar 28 '22 at 09:46
Is it that now we can say any combination kf primes numbers will also not work hence it will cover all possible cases @richrow – ProblemDestroyer Mar 28 '22 at 15:28
Yes, because if there isn't any prime that divides the gcd, any composite number won't work because it is always divided by some prime. Therefore, 1 is the only possible solution. – xImperiak Mar 28 '22 at 16:16
Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 28 '22 at 23:16
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@BillDubuque To be honest, I don't think that this question is a duplicate. From my point of view, it is rather an application of the fact "if $(a,b)=(a,c)=1$, then $(a,bc)=1$" with some technical details (and the author gave some ideas in that direction). Of course, answers to the aforementioned question might be useful, but they don't provide a clear way how to solve this problem (the most useful answer is the IttayWeiss' one). – richrow Mar 29 '22 at 13:25

Calculate $\gcd(a^2b^2, a^2 + ab + b^2)$

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