solve: y^3 = x^3 - 8x^2 + 6x + 8 for x,y are +ve integers?

Question

requires analytical mind

i want all the answers therefore also prove that it does not contain more roots

if u have guts then solve it

Answer 1

Yes, this is a difficult one.
I'll keep trying, but don't hold your breath.
All I can say so far is that I could find
only one solution : (x, y) = (7,1).
There do not seem to be any more
solutions with x >= 8, and this is what
needs to be proved. So good luck with
finding someone who can do this.

Answer 2

a number of you human beings have not got the main outstanding to be so rude. try being greater polite. that's a sturdy question. observe that (x + 8/3)^3 = x^3+8x^2+(sixty 4/3)x+512/27, So y^3 = (x + 8/3)^3 - (80 two/3)x - 296/27 = (x + 8/3)^3 * (a million - ((80 two/3)x-296/27)/(x+8/3)^3 ) The latter area gets small for large x, so for increasing large constructive x, y techniques y ~ (x + 8/3), or x + 2.667. The lesson right that's that y is going to stay exceptionally on the area of x. we can consequently basically evaluate all situations like y = x + a million, or y = x + 2, or y = x + 3, etc., because of the fact there's no longer many to think approximately (y and x, as integers, can basically fluctuate by an integer, and that's easy to tutor (which you may do) that if x and y are the two non-unfavourable, then y > x.) enable's evaluate each and every of those to our equation. (x+a million)^3 = x^3+3x^2+3x+a million (x+2)^3 = x^3+6x^2+12x+8 (x+3)^3 = x^3+9x^2+27x+27 it is ordinary to tutor (which you may do) that this final one is often larger than x^3+8x^2-6x+8 for non-unfavourable x, and oftentimes that (x+h)^3 = x^3+3hx^2+3h^2x+h^3 is often larger than x^3+8x^2-6x+8 while h is an integer 3 or larger. So those can on no account be equivalent - we will not get solutions the place y exceeds x by greater effective than 2. So this leaves the only opportunities as y = x+a million y = x+2 Plug those into the unique equation, cancel words, and you get a quadratic equation in x for each. The y=x+a million case produces non-actual roots for x; the y=x+2 case produces 0 and 9 for x. So, the only available ideas are x=0, y=2 x=9, y=11

Answer 3

To quote the late Professor Mordell: "This is
a very, very difficult question indeed".
Unless you know a clever way to do this,
the solution requires an analysis of the
cubic field Q(x) = x^3 - 8x^2 + 6x + 8.
I ran this through PARI and found that
it is a totally real cubic field with trivial
class group and 2 fundamental units.
I'm sure I could reduce to solving
a collection of Thue equations, but,
unless this question is related to some
other problem, why spend all that time
on it?
Incidentally, BJ, you forgot that there are
2 unknowns here. By taking y large
enough, your argument falls apart. Sorry!

Answer 4

x=7 ,y=1 since by seeing in equation it has 2 sign change that it must hav two negative roots.so only one positive root remains i.e ans

Answer 5

y^3=x^3-8x^2+6x+8
put x=0
y^3=8
so y=2
put y=0
x=-0.6 or 1.6
so the solution set
(0,2),(-0.6,0),(1.6,0)