Find all integer for triplets$(a,b,c)$ [detail]

Question

Find all integer for triplets$(a,b,c)$ that

$a+b+c =3$

$a+b^2+c^2=17$

and $a^2 +b^3+c^3 =21$

I have try to start with

$17-a = ( 3-a )^{2} - 2bc \ \ $ and $21-a^2 = (3-a)\left ((17-a) - bc \right ) +3abc$

but It seems to not be working for me. Could you please guide me any brilliant solution for this problem? Thank you very much.

@Dietrich, note that the current question asks about $a+b^2+c^2$, not $a^2+b^2+c^2$, and about $a^2+b^3+c^3$, not $a^3+b^3+c^3$. — Gerry Myerson, Jan 27 '19 at 11:21
@GerryMyerson Thank you, corrected. It looked so similar to this post, I am sorry. — Dietrich Burde, Jan 27 '19 at 11:25

score 2 · Answer 1 · answered Jan 27 '19 at 11:29

2

From the first two equations, you get $14=(b^2-b)+(c^2-c)$, from which you get $58=(2b-1)^2+(2c-1)^2$. Can you take it from there?

answered Jan 27 '19 at 11:29

Gerry Myerson

how do you get to the last equation you listed in your answer? – poetasis Jan 27 '19 at 12:35
Guessing that factorize out (e.g. 4b^2 - 4b + 1) which combine every single term from first two question. – ABCDEFG user157844 Jan 27 '19 at 15:40
1

@poe, you multiply both sides by 4, add 2 to both sides, and factor. – Gerry Myerson Jan 27 '19 at 21:03

1 Answers1