What is the factorization of x^4-2x^3+3x^2-2x+1?

[cupcakes]

Homework Statement

Given:
x⁷ + y⁷ =1
x + y = 1

Find the integer value(s) of (x-y)².

Homework Equations

The Attempt at a Solution

I thought of substituting for y and then finding the rational roots but then I realized x and y don't have to be rational numbers for (x-y)^2 to be an integer. I am 90% sure the only solution is 1 (when x=1,y=0 or x=0,y=1) but don't know how to prove it. Any hint? Thanks.

 

[SammyS]

Homework Statement

Given:
x⁷ + y⁷ =1
x + y = 1

Find the integer value(s) of (x-y)².

Homework Equations

The Attempt at a Solution

I thought of substituting for y and then finding the rational roots but then I realized x and y don't have to be rational numbers for (x-y)^2 to be an integer. I am 90% sure the only solution is 1 (when x=1,y=0 or x=0,y=1) but don't know how to prove it. Any hint? Thanks.

What are the solutions to the system of equations:

x⁷ + y⁷ =1

x + y = 1

?

 

[cupcakes]

I think I'm very close to solving it. I substituted y= (1-x) in the first equation and expanded. Then I factored that. However I don't know how to factor the last term.

x⁴-2x³+3x²-2x+1

WolframAplha say it can be factored into (x²-x+1)². I just need to figure out how to factor that into this and then I'm done (I think). Any ideas? Thanks.

 

[Mentallic]

I think I'm very close to solving it. I substituted y= (1-x) in the first equation and expanded. Then I factored that. However I don't know how to factor the last term.

x⁴-2x³+3x²-2x+1

WolframAplha say it can be factored into (x²-x+1)². I just need to figure out how to factor that into this and then I'm done (I think). Any ideas? Thanks.

How did you get to that expression?
Expanding (1-x)⁷ allows us to cancel the 1 and -x⁷ terms, so the highest power of x is 6, then after dividing through by x it should be a max power of 5.

 

[cupcakes]

How did you get to that expression?
Expanding (1-x)⁷ allows us to cancel the 1 and -x⁷ terms, so the highest power of x is 6, then after dividing through by x it should be a max power of 5.

Exactly. After expanding and canceling out we have:

7x⁶ - 21x⁵ + 35x⁴ - 35x³ + 21x² - 7^x.

First I divided by (7x). Then I realized (x-1) is a factor (since x=1 is a zero). After long division I have:

(7x)(x-1)(x⁴ - 2x³ + 3x² - 2^x + 1)

 

[Mentallic]

Ahh ok so you already factored out the x=1 factor.

So it's a quartic and hence it must have 4 complex roots. But since all the coefficients are real, the complex roots must come in complex conjugate pairs, and when you expand out $(x-\alpha)(x-\beta)$ where $\alpha, \beta \inℂ$ it must be equal to a quadratic with real coefficients.

So, with this we can deduce that the quartic must be able to be factorized into

$$(x^2+ax\pm 1)(x^2+bx\pm 1)$$

And expanding that, then equating coefficients we can deduce a=b=-1 and we need to take the positive of the $/pm$ operator.

 

[Ray Vickson]

Homework Statement

Given:
x⁷ + y⁷ =1
x + y = 1

Find the integer value(s) of (x-y)².

Homework Equations

The Attempt at a Solution

I thought of substituting for y and then finding the rational roots but then I realized x and y don't have to be rational numbers for (x-y)^2 to be an integer. I am 90% sure the only solution is 1 (when x=1,y=0 or x=0,y=1) but don't know how to prove it. Any hint? Thanks.

One way to proceed is to look for the intersection of the two graphs y = 1 - x and y = (1-x^7)^(1/7). For the latter: look at the graph y vs x for x^n + y^n = 1. When n = 1 you get the line x+y=1. When n = 2 you get the circle x^2 + y^2 = 1, a circle of radius 1 passing through the points (1,0) and (0,1). What happens if n > 2? Well, any point (x,y) in the interior of the first quadrant and on the circle x^2 + y^2 = 1 must lie to the left and below the curve y = f(x) for x^n + y^n = 1. This is because such a point on the circle has 0 < x < 1 and 0 < y < 1, so x^n < x^2 and y^n < y^2, hence x^n + y^n < 1. That means we need to increase x and/or y to bring the quantity x^n + y^n up to 1. In other words, for n > 2 the graph is outside the circle except at the ends (1,0) and (0,1). That means that the intersection of the graph with x + y = 1 is easy to ascertain.

RGV

 

[SammyS]

I think I'm very close to solving it. I substituted y= (1-x) in the first equation and expanded. Then I factored that. However I don't know how to factor the last term.

x⁴-2x³+3x²-2x+1

WolframAplha say it can be factored into (x²-x+1)². I just need to figure out how to factor that into this and then I'm done (I think). Any ideas? Thanks.

Split the 3x² into x² + 2x² .

x⁴-2x³+3x²-2x+1

=x⁴-2x³+x² + 2x²-2x+1

=(x²-x)² + 2(x²-x) + 1

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