Question

Solve $x^4+y^4=82$ and $x-y=2....\left(2\right)$

Answer 1

$x-y=2$

$\Rightarrow (x-y{)}^2=4$

$\Rightarrow x^2+y^2=4+2xy$

$\Rightarrow (x^2+y^2{)}^2=(4+2xy{)}^2$

$\Rightarrow x^4+y^4+2x^2{y}^2=16+4x^2{y}^2+16xy$

$\Rightarrow 82=16+16xy+2x^2{y}^2$

$\Rightarrow (xy{)}^2+8xy-33=0$

$\Rightarrow (xy{)}^2+11xy-3xy-33=0$

$\Rightarrow xy=3\text{or}xy=-11$

If $x-y=2\Rightarrow x=y+2$

For $xy=3\Rightarrow y(y+2)=3\Rightarrow y=1\text{or}y=-3$

Thus, one set of solution is $(3,1)\text{and}(-1,-3)$

Similarly, for $xy=-11\Rightarrow y(y+2)=-11$

This will have no real roots as discriminant is less than zero.

Hence, the solutions are $(3,1)\text{and}(-1,-3)$