# When does (x+y)^4 =

1. Nov 22, 2004

### JasonRox

$$(x+y)^4=x^4+y^4$$

We know this is true if x and/or y = 0, but what other possibilities are there?

I solved for (x+y)^2=x^2+y^2, which has x=-y as a possibility.

The solution isn't necessary, a couple tips should lead me in the right direction.

I broke it down in several ways, but a hint for something to look for would be great. I happen to always be stuck with something along the line like...

STUFF+x^2y^2

I can get the stuff area to equal to zero, but because the are both odd or even numbers that are equal I am always stuff with the odd one at the end.

I lost my paper that shows what STUFF is (stupid I know), but I'll work out a second time and maybe then I'll find a solution.

Thanks.

2. Nov 22, 2004

### Gokul43201

Staff Emeritus
Expand the LHS, and cancel off the common terms with the RHS. You have 3 remaining terms =0. Factoring out xy (stipulating that they be not zero) from these gives you a quadratic, which has no real solutions, and this, you'll figure when you get to it...

3. Nov 22, 2004

### TenaliRaman

I solved for (x+y)^2=x^2+y^2, which has x=-y as a possibility.

Are u sure?
Substitute x=-y on the LHS and then in the RHS .. are they equal?
Simple analysis ,
(x+y)^2 = x^2+2xy+y^2
if (x+y)^2=x^2+y^2
implies
x^2+2xy+y^2 = x^2+y^2
implies
2xy = 0
implies
x or y or both = 0

A similar analysis has been outlined by Gokul in his last post for the power 4 case.

-- AI

4. Nov 22, 2004

### HallsofIvy

(x+y)4= x4+ 4x3y+6x2y2+4xy3+ y4.

Setting that equal to x4+ y4, we can cancel the x4 and y4 terms on both sides leaving

4x3y+ 6x2y2+ 4xy3= 0 or
2xy(2x2+ 3xy+ 2y2)= 0

Clearly x= y= 0 works. It should be evident that 2x2+ 3xy+ 2y2= 0 has no real solutions so x= y= 0 is again the only solution.

5. Nov 22, 2004

### JasonRox

Nevermind about x=-y. I was looking at the wrong solution.

Last edited: Nov 22, 2004