<s>[INST] Spivak 's Calculus (Chapter I, Problem 16c) In problem 16(c) of chapter 1 of Calculus, Spivak asks the reader to determine the conditions under which the expression $(x + y)^4$ equals $x^4 + y^4$. Clearly,
$$ (x + y)^4 = x^4 + y^4 \Leftrightarrow x = 0 \vee y = 0 \vee 4x^2 + 6xy + 4y^2 = 0 $$
From the preceding problem, we know that $0 \leq 4x^2 + 6xy + 4y^2$. If $x = 0$ and $y = 0$ then $4x^2 + 6xy + 4y^2 = 0$. If either $x = 0$ and $y \neq 0$ or $x \neq 0$ and $y = 0$ then $0 < 4x^2 + 6xy + 4y^2$. I want to show that if $x \neq 0$ and $y \neq 0$ then $0 < 4x^2 + 6xy + 4y^2$, which is intuitively true. In order to show that $0 < 4x^2 + 6xy + 4y^2$, it suffices to show that $6xy < 4x^2 + 4y^2$, but I'm not sure how to demonstrate that this inequality is true. I would presumably derive it from the ordered field axioms in conjunction with the local assumptions of the problem.
 [/INST] You have $(x+y)^4 - (x^4 + y^4) = 2 x y (2 y^2 + 3xy + 2 y^2)$.
To find the zeros of $2 y^2 + 3xy + 2 y^2$, use the quadratic formula to get $ y = \frac{x}{4} ( -3 \pm i \sqrt{7})$, hence the only zero (in $\mathbb{R}^2$) is $(0,0)$.
It follows that the zeros are $x=0$ or $y=0$.
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