If roots of $a(x^2+m^2)+amx+cm^2x^2=0$ are $u,v$

Question

If roots of $$a(x^2+m^2)+amx+cm^2x^2=0.......(1)$$ are $u,v$, one can easily prove that $$a(u^2+v^2)+auv+cu^2v^2=0.........(2)$$

Interestingly, if (1) is $f(x,m)=0$, then (2) is $f(u,v)=0.$

The question: Why is this happening?

Answer 1

Let $f(x,y) = a(x^2 + y^2) +ayx + cy^2x^2 $

For a given $m$, we are told that there exists $u, v$ such that $f(u, m) = 0 $ and $f(v, m) = 0 $.
OP claims that $f(u, v) = 0 $, and that there is an easy thought unstated proof.
Presumably they are asking if we can shed light on any deeper meaning.

Treating this as a quadratic in $x$ with $ y = m$ fixed, the 2 roots satisfy

• $u+v = \frac{ -am}{a+cm^2} $.
• $uv = \frac{ am^2 } { a+cm^2 } $.
• In particular, $ 1/u + 1/v + 1/m = 0 $ is a necessary condition. Given 2 values, this uniquely determines the third, so it's also a sufficient condition. To me, this is the "deeper meaning", and you will see how it is used.

We now exploit the symmetry of $x, y$.

Treating this as a quadratic in $y$ with $x = u$ fixed, suppose that $0 = f(u, m) = f(u, n)$ (We know that $n$ exists, eg from Vieta), then

• $m+n = \frac{ -au}{a+cu^2 } $
• $mn = \frac{ au^2 } { a+cu^2 }$
• As before $ 1/m + 1/n + 1/u = 0 $ is a necessary and sufficient condition.

Thus, $ n = v$, so $f(u, n) = 0 $.

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Note

• OP mentioned that this doesn't work for $f(x,y) = a(x^2+y^2) + bxy + cx^2y^2$, which I agree with. Do you see where in this proof it breaks down?