MHB Find relation equation between a,b

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The discussion focuses on finding a relationship between the constants a and b in the equations y = x^3 - ax^2 - bx and y = ax + b, where x is a negative integer and y is a positive integer. Participants explore how to derive this relationship and identify pairs of (x, y) that satisfy both equations. The conversation also touches on the co-primality of x^3 and 1+x, with references to Euclid's algorithm for clarification. The primary goal is to establish the conditions under which the equations hold true for the specified integer constraints. Understanding these relationships is crucial for solving the equations effectively.
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$y=x^3-ax^2-bx---(1)$
$y=ax+b---(2)$
$a,b\in R$
$x$ is a negative integer, $y$ positive integer
(1) find the relative equation between $a$ and $b$
(2)pair(s) of $(x,y)$
 
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Albert said:
$y=x^3-ax^2-bx---(1)$
$y=ax+b---(2)$
$a,b\in R$
$x$ is a negative integer, $y$ positive integer
(1) find the relative equation between $a$ and $b$
(2)pair(s) of $(x,y)$

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

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 $x +1 = \pm 1$
 
kaliprasad said:
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
 
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