请求帮助求解方程组$$r^2+s^2=u^2+v^2$$与$$r^3+s^3=u^3+v^3$$的实互异解

Hey there! Let's break down how to find the distinct real solutions to this system:
$$r^2+s2=u^2+v2$$
$$r^3+s3=u^3+v3$$

First, we'll use algebraic identities to simplify things. Remember these key formulas:

• $r^2 + s^2 = (r+s)^2 - 2rs$
• $r^3 + s^3 = (r+s)(r^2 - rs + s^2) = (r+s)\left[(r+s)^2 - 3rs\right]$

Step 1: Substitute to Simplify the System #

Let’s define substitution variables to make the equations easier to handle:

• $A = r + s$, $B = rs$ (sum and product of $r$ and $s$)
• $C = u + v$, $D = uv$ (sum and product of $u$ and $v$)

Rewriting the original equations with these substitutions gives us:

1. $A^2 - 2B = C^2 - 2D$ → $(A-C)(A+C) = 2(B-D)$ --- (1)
2. $A(A^2 - 3B) = C(C^2 - 3D)$ → $(A-C)(A^2 + AC + C^2) = 3(AB - CD)$ --- (2) (after factoring the difference of cubes on the left)

Step 2: Split into Cases #

We have two main cases to consider based on whether $A = C$ or $A ≠ C$.

Case 1: $A = C$ (r + s = u + v) #

If the sums are equal, plugging into equation (1) gives $B = D$ (rs = uv). This means $(r,s)$ and $(u,v)$ are roots of the same quadratic equation—so they're either identical or permutations of each other. Since we need distinct solutions, we can discard this case entirely.

Case 2: $A ≠ C$ (r + s ≠ u + v) #

Since $A ≠ C$, we can divide both sides of equation (2) by $(A-C)$:
$$A^2 + AC + C^2 = 3\left( \frac{AB - CD}{A - C} \right)$$

Now rearrange equation (1) to solve for $B$:
$$B = D + \frac{(A-C)(A+C)}{2}$$

Substitute this $B$ into the simplified equation (2), then expand and simplify to solve for $D$ and $B$:

• $D = \frac{-(A + 2C)(A - C)}{6}$
• $B = \frac{(A - C)(2A + C)}{6}$

Step 3: Ensure Real Roots #

For $r,s,u,v$ to be real numbers, the quadratics defining them must have non-negative discriminants:

• For $(r,s)$: $A^2 - 4B ≥ 0$ → simplifies to $A^2 - 2AC - 2C^2 ≤ 0$
• For $(u,v)$: $C^2 - 4D ≥ 0$ → simplifies to $2A^2 + 2AC - C^2 ≥ 0$

Combining these inequalities gives us constraints on the ratio $k = \frac{A}{C}$:

• If $C > 0$: $\frac{-1 + \sqrt{3}}{2} ≤ k ≤ 1 + \sqrt{3}$ (excluding $k=1$, since $A≠C$)
• If $C < 0$: $1 - \sqrt{3} ≤ k ≤ \frac{-1 - \sqrt{3}}{2}$ (excluding $k=1$)

Step 4: Explicit Distinct Real Solutions #

To get concrete solutions, pick any $A$ and $C$ that satisfy the above constraints (with $A≠C$). For example:
Let $C=1$ and $A=2$ (which is in the valid interval):

• $B = \frac{(2-1)(4+1)}{6} = \frac{5}{6}$, so $r$ and $s$ are roots of $x^2 - 2x + \frac{5}{6} = 0$ → $r = 1 + \frac{\sqrt{6}}{6}$, $s = 1 - \frac{\sqrt{6}}{6}$
• $D = \frac{-(2+2)(2-1)}{6} = -\frac{2}{3}$, so $u$ and $v$ are roots of $x^2 - x - \frac{2}{3} = 0$ → $u = \frac{1 + \sqrt{\frac{11}{3}}}{2}$, $v = \frac{1 - \sqrt{\frac{11}{3}}}{2}$

You can verify these values satisfy both original equations, and all four numbers are distinct.

In general, for any valid ratio $k = \frac{A}{C}$, you can generate distinct real solutions by calculating $B$ and $D$ as above, then solving the quadratics for $r,s,u,v$.

内容的提问来源于stack exchange，提问作者user529392