Determine the real number $A$ such that the following equation has a real answer for $C$ for every real $B $ : A² + B² + C² + 2ABC = 1

Question

Determine the real number $A$ such that the following equation has a real answer for $C$ for every real $B$ :

A² + B² + C² + 2ABC = 1

first I tried numbering B and find some good equality for example if we choose B=1 we can find A=C but surely this is does not answer the problem could you guys help to solve this problem?

Answer 1

Solve for $C$:

$$C=-AB\pm\sqrt{A^2B^2-A^2-B^2+1}$$

We need a real $A$ such that $A^2B^2-A^2-B^2+1\geq 0$ for any real $B$. Notice that we can factor this to get

$$(A^2-1)(B^2-1)\geq0.$$

Since $B^2-1\geq0$ for all $|B|\geq1$ and $B^2-1<0$ for all $|B|<1$, there are only two $A$-values that can maintain the above inequality regardless of the $B$-value; that is, $A=\pm1$.

Answer 2

Whichever variable we solve for, we discover the same thing.

\begin{align*} A² &+ B² + C² + 2ABC = 1\\ A² &+ (2BC)A + (B² + C² - 1) = 0\\ \\ A &= \frac{-2BC\pm\sqrt{(2BC)^2-4(1)(B² + C² - 1)}}{2}&\\ A &= -BC\pm\sqrt{(B^2 - 1) (C^2 - 1)}\\ \end{align*}

Here, we can see that both variables under that radical must be greater than one of less than one. Since the variables are interchangeable, we know that all three must share this property so, either $A,B,C \le 1$ or $A,B,C \ge1,\space$ –– in absolute value that is.