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Here is an other random conjecture which I have no clue how to prove:

$a,b\in\mathbb N^+\wedge a^2+b^2+ab\in\mathbb P\implies\exists$ $A,B\in\mathbb N^+:A^2+B^2-AB=a^2+b^2+ab$.

Tested for $a,b<20,000$ on my 32 bit tabloid.

I would like to see a proof or an counter-example.

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    $\begingroup$ Did you notice any patterns? What happens if you put $A=a, B=a+b$ $\endgroup$ – Mark Bennet Sep 2 '17 at 18:02
  • $\begingroup$ What does it matter that it's prime? $\endgroup$ – Randall Sep 2 '17 at 18:04
  • $\begingroup$ @Randall: I must have done something wrong, because it seemed only work for primes. $\endgroup$ – Lehs Sep 2 '17 at 18:11
  • $\begingroup$ @Mark The "pattern" has to do with reflections on conics - see my answer. This is closely related to Vieta jumping and similar topics. $\endgroup$ – Bill Dubuque Sep 2 '17 at 18:30
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$$(a+b)^2-(a+b)b+b^2=a^2+ab+b^2.$$

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Hint $ $ If $\,x_1=a\,$ is a root of $\,f(x)\ =\ x^2+\,\color{#c00}b\,x+b^2\,$ then so too is $\,\overbrace{x_2 = -\color{#c00}b-x_1}^{\large {\rm root\ sum}\ =\ -\color{#c00}b} = -b-a$

So $\,-x_2 = a\!+\!b\,$ is a root of $\,f(-x) = x^2-b\,x+b^2$

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