# 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.

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That's because I stated the problem incorrectly. It should have read $(x + y)^4 = x^4 + y^4$. Sorry about that. –  danportin Jul 28 '12 at 5:59
quadratic formula in x (or y)... –  Chris Gerig Jul 28 '12 at 6:13

Use the identity $$4x^2+6xy+4y^2=x^2+y^2+(x+y)^2+(x+y)^2+(x+y)^2.$$

Any square $w^2$ is $\ge 0$, with equality iff $w=0$. The sum of objects that are $\ge 0$ is $\ge 0$, with equality only when all the objects are $0$. This forces $x=y=0$.

Remark: The approach above is minimalist in that we use only facts true in all ordered fields. If we are willing to use properties such as existence of square roots of positive numbers, then we can complete the square in the traditional way.

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This makes sense. So from $0 < (x + y)^2$, $0 < x^2$ and $0 < y^2$ we have $0 < x^2 + y^2 + 3(x + y)^2$. So the only case where $4x^2 + 6xy + 4y^2 = 0$ is $x = 0$ and $y = 0$. –  danportin Jul 28 '12 at 6:24
Run the argument the other way. From $4x^2+6xy+4y^2=0$ conclude that $x=0$ and $y=0$ (and, superfluously, $x+y=0$). So the possibilities (from the original $4x^3y+6x^2y^2+4xy^3=0$) are $x=0$ or $y=0$ or both are $0$, which is included in the or. –  André Nicolas Jul 28 '12 at 6:33
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 $y=0$.