We have to solve the simultaneous equations

x^2 + y^2=10 ...(1)

x^4 + y^4=82 ...(2)

let a = x^2 and b = y^2

This gives

a + b = 10 and a^2 + b^2 = 82

a + b = 10

=> a^2 + b^2 + 2ab = 100

substitute a^2 + b^2 = 82

=> 82 + 2ab = 100

=> 2ab = 18

=> ab = 9

substitute a = 10 - b

b(10 - b) = 9

=> b^2 - 10b = -9

=> b^2 - 9b - b + 9 = 0

=> b(b - 9) - 1(b - 9) = 0

=> (b - 1)(b - 9) = 0

=> b = 1 and b = 9

For b = 1, a = 9

for b = 9, a = 1

Now b = y^2 = 1 => b = -1, + 1 and a = x^2 = 3 , -3

y^2 = 9 => y = -3 and +3 and x^2 = 1 => x = -1 and +1

**The required solutions are: (3, -1), (3 , 1), (-3 , -1), (-3, 1), (1 , -3), (1 , 3) ,(-1 , 3) and (-1 , -3)**

x^2+y^2=10 (1)

x^4+y^4=82 (2)

(x^2+y^2)^2= x^4+2x^2y^2+y^4

Since x^2+y^2=10 and x^4+y^4=82 we have:

100=82+2xy^2

then:

2x^2y^2=18

(xy)^2=9

xy=3 xy=-3

from (1)

x^2+y^2+2xy= 10+6=16

(x+y)^2=16 x+y=4 xy=3

being symmetric: x=1 y=3 x=3 y=1

x+y=-4 xy=3 x=-1 y=-3 x=-3 y=-1

or

x^2+2xy+y^2= 10-6=4

(x+y)^2=4 x+y=2 xy=-3 x=3 y=-1 x=-1 y=3

or

(x+y)^2=-4 x+y=-2 xy=-3 x=-3 y=1 x=1 y=-3

So the couples we are searching for are:

**(1,3), (3,1),(-3,-1),(-1,-3),(3,-1),(-1,3),(-3,1),(1,-3) **

Since the 2nd equation could be written as:

x^4 + y^4 = (x^2 + y^2)^2 - 2(xy)^2

But x^2 + y^2 = 10 (from the 1st equation),

x^4 + y^4 = 10^2 - 2(xy)^2

82 = 100 - 2(xy)^2

2(xy)^2 = 100 - 82

2(xy)^2 = 18

(xy)^2 = 9

xy = +3 or xy = -3

We'll re-write the 1st equation:

x^2 + y^2 = (x+y)^2 - 2xy

10 = (x+y)^2 - 2xy

For xy = 3, we'll get:

10 = (x+y)^2 - 6

(x+y)^2 = 16

(x+y) = 4 or (x+y) = -4

We'll create the quadratic when we know the sum and the product of the roots:

x^2 - Sx + P = 0

For s = x+y = 4 and P = 3

x^2 - 4x + 3 = 0

x1 = [4 + sqrt(16 - 12)]/2

x1 = (4+2)/2

x = 3 and y = 1

x = 1 and y = 3

For s = x+y = -4 and P = 3

x^2 + 4x + 3 = 0

x1 = [-4 + sqrt(16 - 12)]/2

x = (-4 + 2)/2

x = -1 and y = -3

x = -3 and y = -1

For xy = -3, we'll get:

10 = (x+y)^2 + 6

(x+y)^2 = 4

x + y = 2 or x + y = -2

For s = x+y = 2 and P = -3

x^2 - 2x - 3 = 0

x = [2 + sqrt(4 + 12)]/2

x = (2+4)/2

x = 3 and y = -1

x = -1 and y = 3

For s = x+y = -2 and P = -3

x^2 + 2x - 3 = 0

x = [-2 + sqrt(4 + 12)]/2

x = (-2 + 4)/2

x = 1 and y = -3

x = -3 and y = 1

**The solutions of the system of equations are: (3;1);(1;3);(-3;-1);(-1;-3);(3;-1);(-1;3);(1;-3);(-3;1).**