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kaliprasad
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Find all solutions in integers $x,y$ of the equation $y^2+2y= x^4+20x^3+104x^2 + 40x + 2003$
kaliprasad said:Find all solutions in integers $x,y$ of the equation $y^2+2y= x^4+20x^3+104x^2 + 40x + 2003-----(1)$
Albert said:let $p=x^2+10x=x(x+10)-----(2)$
from (1) we have :$y(y+2)=p^2+4p+2003$
$\therefore (y+1)^2=p^2+4p+2004---(3)$
from $(2):x^2+10x-p=0---(4)$
for $x,y$ both are intgers we get :$p>0,$ and $p$ may take values from the following lists:
$11(1\times 11),24(2\times 12),39(3\times 13),----,119(7\times 17)----,n\times (n+10)$
from $(3) : p^2+4p+2004$ is a perfect square
we may assume $p^2+4p+2004=(p+10)^2\rightarrow p=119$
hence $(x,y)=(7,128) \, (7,-130)$
or $(x,y)=(-17,128) \, (-17,-130)$
kaliprasad said:Though the ans is right but
we may assume $p^2+4p+2004=(p+10)^2\rightarrow p=119$
is a weak assumption and some solutions could have been missing
An integer is a whole number that can be positive, negative, or zero.
To find x and y in an equation, you must solve for each variable by manipulating the equation using algebraic operations such as addition, subtraction, multiplication, and division.
The purpose of finding x and y in this equation is to determine the values of x and y that satisfy the equation and make it true. This is important in understanding and solving mathematical problems.
The numbers in the equation represent coefficients and constants that affect the values of x and y. These numbers determine the shape and position of the graph of the equation.
No, there are various methods to solve this equation. One way is to use the quadratic formula, while another is to graph the equation and find the intersection points. Additionally, you can use numerical methods such as iteration or approximation to find the solutions.