**x**

^{2}+ y^{2}= z^{2}(1) We know that we can assume that

**x,y,z**are

*coprime*. [See my previous blog for details]

(2) The second important insight is that

**z**has to be odd.

(a) Assume the opposite that

**z**is even.

(b) Then, there exists another value

**Z**such that

**z = 2 * Z**

(c) Also,

**z**is then divisible by 4 since:

^{2}**z**

^{2}= (2 * Z)^{2}= 4 * (Z^{2})(d) We know that

**x,y**must both be odd because of (1).

(e) Since they are odd, there must also exist values

**X,Y**such that:

**x = 2 * X + 1**

**y = 2 * Y + 1**

(f) But

**x**cannot be divisible by

^{2}+ y^{2}**4**since:

**x**

= 4X

= 4[ X

^{2}+ y^{2}= (2 * X + 1)^{2}+ (2 * Y + 1)^{2}== 4X

^{2}+ 4X + 1 + 4Y^{2}+ 4Y + 1 == 4[ X

^{2}+ X + Y^{2}+ Y ] + 2(g) So, we have a contradiction and we reject our assumption.

(3) Since

**z**is odd, either

**x**or

**y**must be even since an odd number is always the sum of an odd and an even number.

(4) Let's assume

**x**is even. The same argument will also work if

**y**is even.

(5) Now, we know that:

**x**

^{2}= z^{2}- y^{2}= (z - y)(z + y)(6) And,

**z - y**and

**z + y**must be even since

**z,y**are odd.

(7) So, we know that there must exist

**u,v,w**such that:

**x = (2u)**

z + y = (2v)

z - y = (2w)

z + y = (2v)

z - y = (2w)

(8) Which means that:

**(2u)**[From (5) and (7)]

^{2}= (2v)(2w)(9) Dividing both sides by

**4**gives us:

**u**

^{2}= v * w(10) We need 1 more insight before the solution. Here it is:

**v,w**are

*coprime*

(a) Assume that v,w are not coprime.

(b) Then, there exists

**d**such that

**d > 1**and

**d**divides both

**v,w**

(c) Then

**d**divides both

**v + w**and

**v - w**

(d) But:

**z + y + z - y =**

2v + 2w

2v + 2w

So

**2z = 2v + 2w**which means that

**z = v + w**

So

**d**divides

**z**

(e) And:

**z + y - (z - y) = 2v - 2w**

So

**2y = 2v - 2w**which means that

**y = v - w**

So

**d**divides

**y**

(f) Which is a contradiction since

**z,y**are

*coprime*[by (1)].

(g) So, we reject our assumption.

(11) By the properties of

*coprimes*, we know from (9),(10) that

**v,w**are themselves squares (see here for proof). [For those who need a review of

*coprimes*, here is a link.]

(12) So, there exists

**p,q**such that:

**v = p**

^{2}**w = q**

^{2}(13) And, we have our solution since:

**z = v + w = p**

^{2}+ q^{2}**y = v - w = p**

^{2}- q^{2}**x = 2u = 2pq**[Since

**u**means

^{2}= vw**u = pq**]

We also know that:

(a) p,q are relatively prime. [Otherwise, z,x,y would not be relatively prime]

(b) p,q are opposite parity (that is, one is odd and one is even) [Since z is odd]

(14) For sure enough:

**(p**

^{2}+ q^{2})^{2}= (2pq)^{2}+ (p^{2}- q^{2})^{2}(15) Now, to generate our answer, we can pick any

**p,q**we want so long as they are integers.

For example, if

**p = 2**and

**q = 1**, we get:

**z = (2)**.

^{2}+ (1)^{2}= 5**y = (2)**.

^{2}- (1)^{2}= 3**x = 2pq = 2*2*1 = 4**

(16) We can do even better than this because we know that for each x,y,z, if they have common factors, the relation still holds.

**z = d[p**

^{2}+ q^{2}]**y = d[p**

^{2}- q^{2}]**x = d[2pq]**

For example, if

**p = 2**and

**q = 1**and

**d = 2**, we get:

**z = (2)(5) = 10**

**y = (2)(3) = 6**

**x = (2)(4) = 8**

And sure enough,

**6**.

^{2}+ 8^{2}= 36 + 64 = 100QED

What's also nice about this result is that it is not too difficult to apply Fermat's method of infinite descent and prove Fermat's Last Theorem for n=4.