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If $a|x|^2 + b|x| + c = a|y|^2 + b|y| +c$, then is it always $|x|=|y|$ ?

Or similarly,

if $a|x|^n + b|x|^{n-1} + c|x|^{n-2} = a|y|^n + b|y|^{n-1} + c|y|^{n-2}$, then is it $|x| = |y|$ ?

More generally, does it happen that,

when, $a|x| + b|y| + c|z| = a|m| + a|n| + c|p|$

[where, a,b,c are constants. Others are variables]

then, $|x|=|m|$ , $|y| = |n|$ and $|z| = |p|$ ?

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    A clear insight can be obtained by graphing the function $f(x) = ax^2+bx+c$. Does $f(x_1)$ ever equal $f(x_2)$ for different arguments $x_1\neq x_2$? If this happens then the equality $x=y$ will not *always* be true. However if $x,y$ are real numbers and the function $f(x)$ is strictly monotone, then it will be true. That is, the continuous real functions that are injective (one-one) are precisely the strictly monotone continuous functions. – 2017-01-21
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    You have made a substantial edit to the Question, although it does not affect the ultimate point that knowing $f(|x|) = f(|y|)$, even in the special cases you set up, does not generally imply $|x|=|y|$. I leave it as an exercise for you to modify the analysis @HenningMakholm gave in his Answer to tackle injectivity on $[0,+\infty)$. – 2017-01-21

4 Answers 4

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In other words, you're asking whether the functions $$ x \mapsto ax^2+bx+c \\ (x,y,z) \mapsto ax+by+cz $$ are injective. No, they are not, in general.

$ax^2+bx+c$ can be injective, namely if (but only if) $a=0$ and $b\ne 0$.

$ax+by+cz$ is never injective no matter what $a$, $b$, and $c$ are.

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No At first case, see $a=0$, $b=0$

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HINT: it follows that $$a(x^n-y^n)+b(x^{n-1}-y^{n-1})+c(x^{n-2}-y^{n-2})=0$$

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    you must factor the expression in the brackets – 2017-01-21
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    I suggest you go into some detail about the factoring and resulting conclusions if you wish your "answer" to be helpful to future Readers. – 2017-01-21
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    we consider the case $$n=3$$ then we have $$(x-y)(a(x^2+y^2+xy)+b(x+y)+c)=0$$ – 2017-01-21
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    ok but in my comment you will find a hint to solve the problem – 2017-01-21
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In the first case, the answer is clear: yes. But not only. You have in general two solutions. Let be: $$s=x-y$$ $$t=x+y$$ $$a(x^2-y^2)+b(x-y)=0$$ $$a(x-y)(x+y)+b(x-y)=0$$

Then $ast+bs=0$, $s(at+b)=0$ That bring us two solutions: $s=0$, $x=y$ and $t=\frac{-b}{a}$, $y=\frac{-b}{a}-x$