An Interesting Polynomial System

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[Blaqjaqshellaq]

For a shortcut at 5:30, you can take t=2*a. Then t^4 + 2*t^2 - 675=(t^2+1)^2 - 676=(t^2+1)^2 -

[shmuelzehavi]

Another method
The original system of equations is:
x+y=1 (1)
x^5+y^5=211 (2)
Eq. (2) may be written as: (x+y)(x^4-x^3 y+x^2 y^2-xy^3+y^4 )=211
Or,
x^4-x^3 y+x^2 y^2-xy^3+y^4=211 (3)
On the other hand,
(x+y)^4=x^4+4x^3 y+6x^2 y^2+4xy^3+y^4=1
By subtracting eq. (3) from the last expression and dividing by 5 we obtain:
x^3 y+x^2 y^2+xy^3=-42
Or,
xy((x+y)^2-xy)=-42
Or.
xy(1-xy)=-42
Or,
(xy)^2-(xy)-42=0 (4)
Eq. (4) is a quadratic in terms of xy which has two real roots: xy=7 and xy=-6.
Therefore, we obtained two cases:
Case 1: x, y are the roots of the quadratic equation: z^2-z-6=0, which are:
z=3 , z=-2
Case 2: x, y are the roots of the quadratic equation: z^2-z+7=0, which are:
z=(1+3i√3)/2 , z=(1-3i√3)/2
Therefore, the set of solutions of he original system of equations is:
{ (x, y) } = { (3, -2), (2, -3), ((1+3i√3)/2, (1-3i√3)/2)), ((1-3i√3)/2, (1+3i√3)/2) }

[SidneiMV]

3⁵ = 243
2⁵ = 32

243 - 32 = 211

(x, y) = {(3, -2); (-2, 3)}

[raghvendrasingh]

Let x= 1/2 +t, y = 1/2- t
1/32+10× 1/8 t^2 +5 ×1/2 t^4 = 211/2
1+40 t^2+80 t^4 = 3376
80 t^4 +40 t^2 - 3375 = 0
16 t^4+8 t^2 - 675 = 0
(4 t^2+1)^2 = 676
4 t^2+1 = 26, - 26
4 t^2 = 25, -27
t = 5/2, - 5/2, 3√3 i/2, - 3√3 i/2
x= 3, - 2, (1+3√3 i)/2, (1 - 3√3 i)/2
y = - 2, 3, (1-3√3 i )/2, (1+3√3 i)/2

[Quest]

3-2=1 & -2+3= 1., ...
(3;-2)(-2;3) Solns