# Two different polynomial expressions for certain primes

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.

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

## 2 Answers

$$(a+b)^2-(a+b)b+b^2=a^2+ab+b^2.$$

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 $\,f(-x) = x^2-b\,x+b^2$