# Is such a polynomial coefficient possible?

Suppose we have a polynomial $$(x+y)^2$$. What I noticed is that if you subtract $$4xy,$$ the result changes to $$(x-y)^2$$. $$(x+y)^2 - 4xy = x^2 + 2xy+ y^2 - 4xy = x^2-2xy+y^2 = (x-y)^2.$$

This has me thinking about the flexibility of changing that sum just by adding or subtraction coefficients of $$xy.$$

So, my question is, does there exist nonzero coefficients $$b, c, d \in \mathbb{C}$$ such that $$b(x+y)^2 + cxy = d(x+y)^2?$$

I don't know why, but for some reason I have this mental roadblock about analyzing multivariate polynomials. The answer seems like "no" by equating coefficients and deriving a contradiction, but on the other hand there's numbers like the golden ratio that satisfy $$1 + \frac{1}{\phi} = \phi$$ and so maybe there's two specific non-zero numbers where such a postulation is possible.

• Check for typos. Did you want any minuses in what you wrote? Otherwise, $b(x+y)^2 + cxy = d(x+y)^2$ is true for $b=d$ and $c=0$ Jan 5, 2022 at 17:32
• I did intend a minus in the first sentence. But is there any non-zero $c$ this is true for? Jan 5, 2022 at 17:33
• Note that two polynomials are equal iff each of their respective coefficients are equal. The coefficient of $x^2$ in the expansion on the left is $b$ and the coefficient of $x^2$ in the expansion on the right is $d$, so these must be equal. Jan 5, 2022 at 17:33
• If you simplify the left-hand side you obtain a $(2b+c)xy$ term, is there anything that can be done with that for non-zero $c$? Jan 5, 2022 at 17:35
• You can edit the question to show us your work equating coefficients and ask if there's any gap in your logic. Your concluding true statement about the golden mean doesn't seem to point to a problem. Jan 5, 2022 at 17:35

$$bx^2 + 2bxy + by^2 + cxy = dx^2 + 2dxy + dy^2.$$
$$bx^2 + (2b+c)xy+by^2 = dx^2 + 2dxy + dy^2.$$
Equating coefficients, one may first find $$b=d,$$ then $$2b+c = 2d \implies c=0.$$