Two different polynomial expressions for certain primes

Question

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 02 '17 at 18:02
@Randall: I must have done something wrong, because it seemed only work for primes. — Lehs, Sep 02 '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 02 '17 at 18:30

score 3 · Accepted Answer · answered Sep 02 '17 at 18:01

3

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

answered Sep 02 '17 at 18:01

Angina Seng

158,341

Better to bring the innate symmetry to the fore rather than pull the answer out of a hat - see my answer (and see reflections on conics and Vieta jumping. – Bill Dubuque Sep 02 '17 at 18:27

Bill Dubuque · Answer 2 · 2017-09-02T18:27:49.283

1

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

edited Sep 02 '17 at 18:27

answered Sep 02 '17 at 18:21

Bill Dubuque

272,048

Two different polynomial expressions for certain primes

2 Answers2