Non-cyclic polynomial with the cyclic permutations equal

Question

Consider $\:\: f(x,y,z) := \alpha^2(x-y) + \alpha(z-y)+ (x-z) \:\:$ with some constant $\alpha \in \mathbb{R} \: \:$ If I'm not wrong, this function isn't cyclic. Now consider $$f(a,b,c) = f(c,a,b) = f(b,c,a) = 0 \:\:\: a,b,c \in \mathbb{R}$$

Does this imply that $a=b=c$?

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Motivation

On attempting the second problem from this contest I employed the following approach $$P_1(\alpha) = P_2(\alpha) \implies \alpha^2(a-b)+\alpha(c-b)+(a-c) = 0 \tag 1 \:\:\:\: \text{(I've omitted some simple algebra)}$$

Similarly we get $$\alpha^2(b-c) + \alpha(a-c) + (c-b) = 0 \tag 2$$ $$\alpha^2(c-a) + \alpha(b-a) + (c-b) = 0 \tag 3$$

After this we define $f(x,y,z)$ and if the property that I am thinking of is true, then I think we're done.

Answer 1

Yes, for real $\alpha$ the only solutions are $a=b=c$.

The system of linear equations $f(a,b,c)=0, f(c,a,b)=0,f(b,c,a)=0$ has coefficient matrix $$ A = \left[ \begin {array}{ccc} {\alpha}^{2}+1&-{\alpha}^{2}-\alpha&\alpha -1\\ -{\alpha}^{2}-\alpha&\alpha-1&{\alpha}^{2}+1 \\ \alpha-1&{\alpha}^{2}+1&-{\alpha}^{2}-\alpha \end {array} \right] $$ which has rank $2$ (for generic values of $\alpha$) and null space spanned by $(1,1,1)^T$. Thus for almost all $\alpha$, the only solutions have $a=b=c$. To have other solutions, you'd need the rank to be at most $1$, which would require all $2 \times 2$ submatrices to have determinant $0$, i.e. all entries of the classical adjoint or adjugate to be $0$. In fact, all entries of the adjugate are $-{\alpha}^{4}-{\alpha}^{3}-2\,{\alpha}^{2}+\alpha-1$. But this polynomial has no real roots.

Answer 2

I think I have a proof for this.

$$(1) + (2) \implies \alpha^2(a-c) + \alpha(a-b) + (b-c) = 0 \tag #$$

$$(2) + (3) \implies \alpha^2(b-a) + \alpha(b-a) + \alpha(b-c) + (c-a) =0$$

Comparing coefficients we have $a-c = b-a$ implying that b,a,c are in arithmetic progression. Thus we replace the differences in # with a common difference $d$

$$\alpha^2(-d) + \alpha(d) -2d = 0 \implies d(-\alpha^2+ \alpha -2)=0$$

So either $d = 0$ or $-\alpha^2+ \alpha -2=0 $ but we know that $-\alpha^2+ \alpha -2 \neq 0$ since $\Delta <0$ and we know $\alpha$ to be real. Therefore $d=0$ and $a=b=c$

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Alternatively $$(1) = (2) \implies a -b = b-c$$ Similarly we have $a-c = c-b$ and $a-c = b-a$

Therefore $a,b,c$ are all the middle term of the same arithemtic progression implying that the common difference is $0$.