Solve the system x⁴ + y⁴ = 82 and xy = 3.

Step-by-step solution

1. 1

   Build a bridge to the lower-degree symmetric quantities. The identity

   $\left(x^{2}+y^{2}\right)^{2}=x^{4}+2x^{2}y^{2}+y^{4}$

   lets the given fourth powers be converted using only $xy$, which is also given. This is the standard route for symmetric systems: climb down from degree $4$ to degree $2$.

2. 2

   Compute x² + y².

   $\left(x^{2}+y^{2}\right)^{2}=82+2(3)^{2}=82+18=100$

   Since $x^{2}+y^{2}\ge 0$ for real $x,y$, only the positive root is admissible:

   $x^{2}+y^{2}=10$

3. 3

   Get x + y and x − y.

   $(x+y)^{2}=x^{2}+y^{2}+2xy=10+6=16\ \Longrightarrow\ x+y=\pm 4$

   $(x-y)^{2}=x^{2}+y^{2}-2xy=10-6=4\ \Longrightarrow\ x-y=\pm 2$

4. 4

   Combine the sign choices, keeping only consistent pairs. From $x+y=4$: with $x-y=2$ we get $(3,1)$; with $x-y=-2$ we get $(1,3)$. From $x+y=-4$: $(-1,-3)$ and $(-3,-1)$.

   $\boxed{(x,y)=(3,1),\ (1,3),\ (-1,-3),\ (-3,-1)}$

5. 5

   Check every pair against both equations. For $(3,1)$: $81+1=82$ ✓ and $3\cdot 1=3$ ✓. For $(-1,-3)$: $1+81=82$ ✓ and $(-1)(-3)=3$ ✓. The other two follow by the symmetry $x\leftrightarrow y$ and $(x,y)\to(-x,-y)$, both of which preserve the system.

6. 6

   Note the sign consistency requirement. Because $xy=3>0$, $x$ and $y$ must have the same sign — which is exactly why $x+y=4$ pairs only with $x-y=\pm 2$ and never produces a mixed-sign solution. Any candidate such as $(3,-1)$ would fail $xy=3$.