Timeline for If $~(c - b) ^ 2 + 3cb = a^3~$ has nonzero integer solutions, then $~(a,c) \gt 1~$ or $~(b,c) \gt 1$?
Current License: CC BY-SA 3.0
7 events
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| Aug 4, 2014 at 17:54 | comment | added | GH from MO | @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 $3$. | |
| Aug 4, 2014 at 8:43 | comment | added | Stefan Kohl♦ | Why do you think so complicated? -- Already $a = 1$, $b = -1$, $c = 1$ is a counterexample! | |
| Aug 4, 2014 at 8:35 | comment | added | Geoff Robinson | 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 $7^{3} = 14^{2}+ (7 \times 14) +7^{2}).$ | |
| Aug 4, 2014 at 3:06 | comment | added | GH from MO | @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). | |
| Aug 4, 2014 at 3:01 | comment | added | Mike | How about if $~c-b~$ is a cubic number? | |
| Aug 4, 2014 at 0:57 | history | edited | GH from MO | CC BY-SA 3.0 |
added 62 characters in body
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| Aug 4, 2014 at 0:50 | history | answered | GH from MO | CC BY-SA 3.0 |