# Blog Archives

## CoPrime Numbers : x^n + y^n = z^n

This post is about the equation:

**x^{n} + y^{n}= z^{n}**

I will show that given any three integers that satisfy this equation, either:

(a) all three of them are coprime with each other

or

(b) it is possible to cancel out common components and derive three numbers that are coprime.

Two numbers are coprime if they do not share any common divisors.

**Definition 1: Coprime**

Two numbers

x,yare said to becoprimeif and only ifddividesx, ddividesy -> d = 1Two numbers

x,yare said to not becoprimeif and only if there exists a valued > 1such thatddividesx, ddividesy

Now, here’s the proof that was promised:

**Lemma: We can reduce any solution to x ^{n} + y^{n} = z^{n} to a form where x,y,z are coprime.**

To prove this, we will need to prove two things:

(1) If a factor divides any two values of this equation, then the n-power of it divides the n-power of the third value.

(2) If an n-power of a factor divides the n-power of a value, then the factor divides the value itself.

Step 1: For x^{n} + y^{n} = z^{n}, the n-power of any common factor of two divides the n-power of the third.

Case I: Let’s assume d divides x, d divides y

(1) There exists x’, y’ such that: x = d(x’), y = d(y’)

(2) z^{n} = x^{n} + y^{n} = (dx’)^{n} + (dy’)^{n}

= d^{n}(x’)^{n} + d^{n}(y’)^{n}

= d^{n}[(x’)^{n} + (y’)^{n}]

Case II: Let’s assume d divides z and d divides x or d divides y

(1) Let’s assume d divides x (the same argument will work for y)

(2) There exists x’, z’ such that: x = d(x’), z = d(z’)

(3) We now say y^{n} = z^{n} – x^{n}

(4) We can now follow the same reasoning as above.

QED

**Step 2: d ^{n} divides x^{n} → d divides x**

(1) Let c be the greatest common denominator (gcd) for **d,x**.

(2) Let **D = d / c, X = x / c**.

(3) Now the gcd of **(X,D) = 1**. [See here for the explanation]

(4) So, the gcd of **(X ^{n},D^{n}) = 1**.

(5) We know that there exists

**k**such that

**x**[Since

^{n}= k * d^{n}**d**divides

^{n}**x**]

^{n}(6) Applying (2), we get

**(cX)**

^{n}= k*(cD)^{n}(7) Which gives us:

**c**

^{n}X^{n}= k * c^{n}D^{n}(8) Dividing

**c**from each side gives:

^{n}**X**

^{n}= D^{n}*k(9) Now it follows that gcd(

**D**) =

^{n},k**1**.

(a) Assume gcd(D

^{n},k) = a, a > 1

(b) Then,adivides D^{n}and X^{n}[From 8]

(c) But gcd(D^{n},X^{n}) ≠ 1.

(d) But this contradicts (4)

(e) So, we reject our assumption.

(10) So, we can conclude that **k** is an n-power. [See blog on Infinite Descent for the proof]

(11) Which means that there exists **u** such that **u ^{n} = k**.

(12) And we get

**D**

^{n}* u^{n}= X^{n}(13) And

**(Du)**

^{n}= X^{n}(14) Implying that

**Du = X**and multiplying by

**c**that

**du=x**.

(15) Which proves that

**d**divides

**x**.

QED