MHB Find relation equation between a,b

[Albert1]

$y=x^3-ax^2-bx---(1)$
$y=ax+b---(2)$
$a,b\in R$
$x$ is a negative integer, $y$ positive integer
(1) find the relative equation between $a$ and $b$
(2)pair(s) of $(x,y)$

 

[kaliprasad]

$y=x^3-ax^2-bx---(1)$
$y=ax+b---(2)$
$a,b\in R$
$x$ is a negative integer, $y$ positive integer
(1) find the relative equation between $a$ and $b$
(2)pair(s) of $(x,y)$

Spoiler

we have $y=x^3- ax^2-bx = x^3 - x(ax+b) = x^3-xy$ using (2)
or $y+xy = x^3$
or $y = \frac{x^3}{1+x}$
now $x^3$ and (1+x) are co-primes and 1 + x is not a factor of $x^3$ unless $1+x = \pm 1$
and $1+x=1$ gives x = 0 which is not admissible
so $1+x = -1$ hence $x = -2 $ and $y=8$ so this is the solution as it satisfied x -ve and y + ve
putting in (2) we get
8 = -2a + b or $2a-b = -8$ is the relationship between a and b

 

[Fermat1]

Spoiler

we have $y=x^3- ax^2-bx = x^3 - x(ax+b) = x^3-xy$ using (2)
or $y+xy = x^3$
or $y = \frac{x^3}{1+x}$
now $x^3$ and (1+x) are co-primes and 1 + x is not a factor of $x^3$ unless $1+x = \pm 1$
and $1+x=1$ gives x = 0 which is not admissible
so $1+x = -1$ hence $x = -2 $ and $y=8$ so this is the solution as it satisfied x -ve and y + ve
putting in (2) we get
8 = -2a + b or $2a-b = 8$ is the relationship between a and b

How do you know $x^3$ and $1+x$ are co-prime?

 

[kaliprasad]

How do you know $x^3$ and $1+x$ are co-prime?

Spoiler

because $x^3+1 = (x+1)(x^2-x + 1)$

so $x^3 = (x+1)(x^2-x+1) + 1$

so $GCD(x^3, x + 1) = GCD(- 1,x +1) =1$ or -1 if you like it and hence co-primes unless $x +1 = \pm 1$

 

[Fermat1]

Spoiler

because $x^3+1 = (x+1)(x^2-x + 1)$

so $x^3 = (x+1)(x^2-x+1) + 1$

so $GCD(x^3, x + 1) = GCD(- 1,x +1) =1$ or -1 if you like it and hence co-primes unless $x +1 = \pm 1$

Euclid's algorithm; ok thanks