Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

2 Answers

up vote 4 down vote accepted

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.

share|improve this answer
    
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
add comment

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 $y=0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.