# Thread: System in Z3

1. ## System in Z3

Find x,y,z in Z solutions for system :

2. $\left\{\begin{array}{ll}x+y=z=12\\x^2+y^2=z^2+12\e nd{array}\right.$

Let $s=x+y, \ p=xy$

Then $\left\{\begin{array}{ll}s=z+12\\s^2-2p=z^2+12\end{array}\right.$

Replace s in the second equation. Then $p=12z+66$. The quadratic equation having the roots such as their sum is s and their product is p is $t^2-st+p=0$

That means $t^2-(z+12)t+12z+66=0$. The roots of this equation are x and y and are integers. Then the discriminant must be a perfect square.

$\Delta=(z-12)^2-264=k^2\Rightarrow(z-k-12)(z+k-12)=264$

$\left\{\begin{array}{ll}z-k-12=2\\z+k-12=132\end{array}\right.\Rightarrow z=79$

Replace z in t-equation: $t^2-91t+1014=0$ with the roots 13 and 78. So, $x=13, \ y=78, \ z=79$ or $x=78, \ y=13, \ z=79$

You can try other cases:

$\left\{\begin{array}{ll}z-k-12=4\\z+k-12=66\end{array}\right.$ or $\left\{\begin{array}{ll}z=k-12=8\\z+k-12=33\end{array}\right.$ etc.

3. Originally Posted by red_dog
$\left\{\begin{array}{ll}x+y=z=12\\x^2+y^2=z^2+12\e nd{array}\right.$

Let $s=x+y, \ p=xy$

Then $\left\{\begin{array}{ll}s=z+12\\s^2-2p=z^2+12\end{array}\right.$

Replace s in the second equation. Then $p=12z+66$. The quadratic equation having the roots such as their sum is s and their product is p is $t^2-st+p=0$

That means $t^2-(z+12)t+12z+66=0$. The roots of this equation are x and y and are integers. Then the discriminant must be a perfect square.

$\Delta=(z-12)^2-264=k^2\Rightarrow(z-k-12)(z+k-12)=264$

$\left\{\begin{array}{ll}z-k-12=2\\z+k-12=132\end{array}\right.\Rightarrow z=79$

Replace z in t-equation: $t^2-91t+1014=0$ with the roots 13 and 78. So, $x=13, \ y=78, \ z=79$ or $x=78, \ y=13, \ z=79$

You can try other cases:

$\left\{\begin{array}{ll}z-k-12=4\\z+k-12=66\end{array}\right.$ or $\left\{\begin{array}{ll}z=k-12=8\\z+k-12=33\end{array}\right.$ etc.
HELLO :THANK YOU
I'have a simpl solution :

Find numbers in Z by divid 66.......(66=2×11×3 )
12 solutions LOOK HERE