Sums of 6th and 7th degree powers

[homegrown898]

This is a very similar question to what I posted earlier.

Basically I am trying to find when (x+y)⁶ = x⁶ + y⁶ assuming that xy≠0

I am trying to play with it algebraically to find a contradiction, but have been unsuccessful

I'm also working on (x+y)⁷ = x⁷ + y⁷ assuming xy≠0

I'm trying to play with it algebraically to show the only other case is when y=-x

 

[Matt Benesi]

Look at the examples over here at Wikipedia's article on the Binomial Theorem.

 

[micromass]

1) We can take y=1. If an (x,y) exists such that the equality holds (with xy nonzero), then there also exists an x' such that the equality holds for (x',1).

2) Take y fixed. The goal is to prove that

f(x)=(x+1)^6-x^6-1^6

has a zero. In order to do this, it suffices to show that f is monotonically increasing. Show this with derivatives.

 

[Matt Benesi]

1) We can take y=1. If an (x,y) exists such that the equality holds (with xy nonzero), then there also exists an x' such that the equality holds for (x',1).

2) Take y fixed. The goal is to prove that

f(x)=(x+1)^6-x^6-1^6

has a zero. In order to do this, it suffices to show that f is monotonically increasing. Show this with derivatives.

f(x)=6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x

Guess where the zero is... anyways, the poster wants a non zero x in which x != -y. Really- remember that (x+y)^n has a simple binomial expansion... and subtracting (x^n+y^n) from it will leave you with the middle part...

 

[micromass]

f(x)=6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x

Guess where the zero is... anyways, the poster wants a non zero x in which x != -y. Really- remember that (x+y)^n has a simple binomial expansion... and subtracting (x^n+y^n) from it will leave you with the middle part...

You're making it a lot harder then it needs to be really.

 

[homegrown898]

1) We can take y=1. If an (x,y) exists such that the equality holds (with xy nonzero), then there also exists an x' such that the equality holds for (x',1).

2) Take y fixed. The goal is to prove that

f(x)=(x+1)^6-x^6-1^6

has a zero. In order to do this, it suffices to show that f is monotonically increasing. Show this with derivatives.

I understand the approach you are taking, but I don't understand how setting y as a constant allows us to show that there is no solution to (x+y)⁶ = x⁶ + y⁶ other than x=0 or y=0 for all possible values of x and y

 

[micromass]

I understand the approach you are taking, but I don't understand how setting y as a constant allows us to show that there is no solution to (x+y)⁶ = x⁶ + y⁶ other than x=0 or y=0 for all possible values of x and y

I just take an arbitrary y. And I try to prove that

f(x)=(x+y)^6-x^6-y^6

has only one zero. That proves it because I took y arbitrary but fixed.