A little long recursion RELOADED

Let $S_1=a+b+c$ , $S_2=ab+ac+bc$ , $S_3=abc$ and $P_n=a^n+b^n+c^n$ .

By Newton's Sums we know:

$P_1=S_1$

$P_2=P_1S_1-2S_2 = S_1^2-2S_2$

$P_3=P_2S_1-P_1S_2+3S_3 = S_1^3-3S_1S_2+3S_3$

$P_n=P_{n-1}S_1-P_{n-2}S_2+P_{n-3}S_3$

Hence, $P_4=P_3S_1-P_2S_2+P_1S_3 = S_1^4-4S_1^2S_2+4S_1S_3+2S_2^2$

With the known values the following system is formed:

$S_1^2-2S_2=25$

$S_1^3-3S_1S_2+3S_3=106$

$S_1^4-4S_1^2S_2+4S_1S_3+2S_2^2=477$

From the first equation, solve for $S_2$ :

$S_2=\dfrac{S_1^2-25}{2}$

Substitute in the second equation, simplify and solve for $S_3$ :

$S_1^3-3S_1\left(\dfrac{S_1^2-25}{2}\right)+3S_3=106 \Longrightarrow S_3=\dfrac{S_1^3-75S_1+212}{6}$

Substitute $S_2$ and $S_3$ in the third equation:

$S_1^4-4S_1^2\left(\dfrac{S_1^2-25}{2}\right)+4S_1\left(\dfrac{S_1^3-75S_1+212}{6}\right)+2\left(\dfrac{S_1^2-25}{2}\right)^2=477$

Expand and simplify:

$S_1^4-150S_1^2+848S_1-987=0$

Using the rational root test it factors as:

$(S_1-7)(S_1^3+7S_1^2-101S_1+141)=0$

The second factor doesn't have natural solutions, hence $S_1=7$ .

Obtain $S_2$ and $S_3$ :

$S_2=\dfrac{S_1^2-25}{2}=\dfrac{7^2-25}{2}=12$

$S_3=\dfrac{S_1^3-75S_1+212}{6}=\dfrac{7^3-75(7)+212}{6}=5$

Now, obtain $P_5$ , $P_6$ and $P_7$ :

$P_n=7P_{n-1}-12P_{n-2}+5P_{n-3}$

$P_5=7P_4-12P_3+5P_2=7(477)-12(106)+5(25)=2192$

$P_6=7P_5-12P_4+5P_3=7(2192)-12(477)+5(106)=10150$

$P_7=7P_6-12P_5+5P_4=7(10150)-12(2192)+5(477)=47131$

Hence:

$x+y+z=\sqrt{2192+17}=47$

$xy+xz+yz=\sqrt{47131-42}=217$

$xyz=\sqrt{10150+51}=101$

Finally, let $P(m)$ have roots $x$ , $y$ and $z$ :

$P(m)=(m-x)(m-y)(m-z)=m^3-(x+y+z)m^2+(xy+xz+yz)m-xyz$

$P(m)=m^3-47m^2+217m-101$

By definition, the discriminant of $P(m)=m^3+bm^2+cm+d$ is:

$\Delta=b^2c^2-4c^3-4b^3d+18bcd-27d^2=(x-y)^2(y-z)^2(z-x)^2$

So, $(x-y)^2(y-z)^2(z-x)^2=(-47)^2(217)^2-4(217)^3-4(-47)^3(-101)+18(-47)(217)(-101)-27(-101)^2=39468212$

And finally:

$\lfloor \sqrt{|(x-y)(y-z)(z-x)|} \rfloor=\lfloor \sqrt[4]{39468212} \rfloor=\boxed{79}$