If $~(c - b) ^ 2 + 3cb = a^3~$ has nonzero integer solutions, then $~(a,c) \gt 1~$ or $~(b,c) \gt 1$?
I think this is true, how to prove this?
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$\begingroup$ maybe you could give a motivation for this question, which I guess is related to your previous one mathoverflow.net/questions/177429 $\endgroup$– Pietro MajerCommented Aug 4, 2014 at 0:32
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$\begingroup$ See also math.stackexchange.com/questions/885181/…. $\endgroup$– Dietrich BurdeCommented Aug 4, 2014 at 7:53
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2 Answers
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It is not true. Let us call an integer nice if its prime divisors are congruent to $1$ modulo $3$. If $n$ is nice, then in the ring of Eisenstein integers it factors as $n=(c-b\omega)(c-b\bar\omega)$, where the factors $c-b\omega$ and $c-b\bar\omega$ are coprime. In particular, $(b,c)=1$ and we have $n=c^2+cb+b^2$. Note that $(b,c)=1$ implies that $n,b,c$ are pairwise coprime.
Now let $a$ be nice, then $n=a^3$ is also nice, hence by the above there exists a representation $$a^3=c^2+cb+b^2=(c-b)^2+3cb$$ such that $a,b,c$ are pairwise coprime.
Here is a concrete counterexample: $a=7$, $b=-18$, $c=19$.
answered Aug 4, 2014 at 0:50
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$\begingroup$ How about if $~c-b~$ is a cubic number? $\endgroup$– MikeCommented Aug 4, 2014 at 3:01
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$\begingroup$ @Mike: $c-b$ cannot be a cube (for distinct nonzero $b$ and $c$) because the Fermat equation $x^3+y^3=z^3$ has no nonzero solution (as proved by Euler). $\endgroup$ Commented Aug 4, 2014 at 3:06
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$\begingroup$ If $n$ is nice, it has at least one such factorization with $c$ and $b$ coprime, (which is all you need for the answer) but if $n$ is not squarefree, then there are such (rational integer) factorizations with $c$ and $b$ not coprime (eg $7^{3} = 14^{2}+ (7 \times 14) +7^{2}).$ $\endgroup$ Commented Aug 4, 2014 at 8:35
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2$\begingroup$ Why do you think so complicated? -- Already $a = 1$, $b = -1$, $c = 1$ is a counterexample! $\endgroup$– Stefan Kohl ♦Commented Aug 4, 2014 at 8:43
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1$\begingroup$ @StefanKohl: You are right. At any rate, having more counterexamples is better. In fact my response gives all the counterexamples $a$ which are not divisible by $3$. $\endgroup$ Commented Aug 4, 2014 at 17:54
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I'm in this thread showed how do the solutions of this equation.
Mutually simple solutions for one of these formulas are coming out.
$$x^2+xy+y^2=z^3$$
$$x=s^3+3ps^2-p^3$$
$$y=p^3+3p^2s-s^3$$
$$z=p^2+ps+s^2$$
If $p=2$ ; $s=3$ We get: $x=73$ ; $y=17$ ; $z=19$
The above example is not correct.
These decisions are determined by a formula.
$$x=s(p^2+ps+s^2)$$
$$y=p(p^2+ps+s^2)$$
$$z=p^2+ps+s^2$$
If $p=19$ ; $s=-18$ ; Then the solutions are. $x=-18(7^3)$ ; $y=19(7^3)$ ; $z=7^3$
That is equivalent to the solution. $x=-18$ ; $y=19$ ; $z=7$
