Find all pair of integers (a,b) satisfying a^3+(a^2)*b+a*(b^2)+b^3=2001
Hello, swarna!
Find all pair of integers $\displaystyle (a,b)$ satisfying: .$\displaystyle a^3+a^2b+ab^2+b^3\:=\:2001$
Factor: .$\displaystyle a^2(a+b) + b^2(a+b) \:=\:2001$
Factor: .$\displaystyle (a+b)(a^2+b^2) \:=\:3\cdot23\cdot29$
There are four possible systems of equations:
. . $\displaystyle \begin{Bmatrix} a+b&=& 1 \\ a^2+b^2&=& 2001 \end{Bmatrix}\quad \begin{Bmatrix}a+b &=& 3 \\ a^2+b^2 &=& 667 \end{Bmatrix} \quad \begin{Bmatrix} a+b &=& 23 \\ a^2+b^2 &=& 87 \end{Bmatrix} \quad \begin{Bmatrix} a+b&=& 29 \\ a^2+b^2 &=& 69 \end{Bmatrix}$
None of them has an integer solution.
Hello, swarna!
From this it can be seen that it has no positive solution.
How did you come to this conclusion?
One system is: .$\displaystyle \begin{Bmatrix}a+b &=& 1 & [2] \\ a^2+b^2 &=& 2001 & [2] \end{Bmatrix}$
And we hope to find a solution in which both $\displaystyle a$ and $\displaystyle b$ are integers.
From [1]: .$\displaystyle b \:=\:1-a$
Substitute into [2]: .$\displaystyle a^2 + (1-a)^2 \:=\:2001$
. . $\displaystyle 2a^2 - 2a - 2000 \:=\:0 \quad\Rightarrow\quad a^2-a-1000 \:=\:0$
Quadratic Formula: .$\displaystyle a \:=\:\frac{1\pm\sqrt{(-1)^2 - 4(-1000)}}{2(1)} \:=\:\frac{1\pm\sqrt{4001}}{2}$
For $\displaystyle a$, we have both a positive solution and a negative solution,
. . but neither of them is an integer.
Somehow, you proved that $\displaystyle a$ is not a positive integer,
. . but it might be a negative integer?
my observations are ...............
for the first case a+b=1, a^2+b^2=2001, clearly it has no positive solution, because sum of two positive integer can not be 1.
for the second case a+b=3 , we get either a=1,b=2 or a=2,b=1. so in either case a^2+b^2=5 but a^2+b^2=667 a contradiction.
for the third case a^2+b^2=87, so the absolute value of both a and b is less than 10 which in turn implies that a+b<20 but given that a+b=23 a contradiction.
for the fourth case a^2+b^2=69,so the absolute value of both a and b is less than 10 which in turn implies that a+b<20 but given that a+b=29 a contradiction.
In this way i conclude that it has no positive solution. Am i wrong??
The product $(a+b)(a^2+b^2)$ is positive if and only if both factors are either positive or negative. The factor $a^2+b^2$ cannot be negative, so $a+b$ cannot be negative. Hence, the solutions (positive or negative) can exist if and only if it satisfies one of the system of equations Soroban posted for you. For the second case:
$a+b = 3, a^2+b^2 = 667$:
Solve for $a$ in the first equation.
$a = 3-b$
Substitute this into the second equation:
$(3-b)^2 + b^2 = 667$
$9 - 6b + b^2 + b^2 = 667$
$b^2 - 3b - 329 = 0$
Using the quadratic equation, you find $b = \dfrac{3 \pm \sqrt{1325}}{2}$, which is not an integer.
For the third system, $a+b = 23$. So, solve for $a$ and plug it into the second equation of that system. You find $b = \dfrac{23 \pm \sqrt{-355}}{2}$, neither value for $b$ is a real number.
For the fourth system, $a+b = 29$. Solve for $a$ and plug it into the second equation of that system. You find $b = \dfrac{29 \pm \sqrt{-703}}{2}$, neither value for $b$ is a real number.
So, just as Soroban said, there are no integer solutions to your equation.