Ignore this post.

Added inEdit:

Pro,

What I meant was: "Ignore my post." I had written something that was incorrect & didn't see how to delete my post.

Results 1 to 5 of 5

- Aug 26th 2011, 12:49 PM #1

- Joined
- Aug 2011
- From
- Omagh
- Posts
- 55

## even more a²+b²=2c²

Back again

For those who didn't see earlier thread I'll just put this first

we can write a, b & c like this:

Proof:

let a=194, b=142 and c=170

using (1)-(2) we have

substituting this in (1), (2) & (3) gives the expected results

All good so far but...

let a=527, b=289, c=425

using (1)-(2) we have

substituting this in (1), (2) & (3) gives

What am I doing wrong or misunderstanding?

Many thanks

Pro

- Aug 26th 2011, 02:11 PM #2

- Joined
- Nov 2010
- From
- Clarksville, ARk
- Posts
- 398

- Aug 26th 2011, 02:19 PM #3

- Joined
- Aug 2011
- From
- Omagh
- Posts
- 55

- Aug 26th 2011, 03:02 PM #4

- Joined
- Nov 2010
- From
- Clarksville, ARk
- Posts
- 398

## Re: even more a²+b²=2c²

Once you choose any

**one**of a, or b, or c, then you have determined the value for m.

Your first example works because m=13 for a = 194, b = 142, and c = 170.

For your second case, if a = 527, then m = 22, so b = 439 c = 485.

If you want b = 289, then , so and

... etc.

You can pick an integer, m, then get a, b, and c, so that but simply finding a, b, c that satisfy does not ensure that there is a number m that generates a, b, and c.

- Aug 26th 2011, 04:52 PM #5

- Joined
- Aug 2011
- From
- Omagh
- Posts
- 55

## [solved] Re: even more a²+b²=2c²

You can pick an integer, m, then get a, b, and c, so that but simply finding a, b, c that satisfy does not ensure that there is a number m that generates a, b, and c.

It just looks**so**right.

The m I gave wasn't just for a=527. It was generated using (1) and (2) as simultaneous equations, but the result doesn't satisfy (1) and (2) simultaneously ¿

Ah! Because the equations aren't simultaneously correct!

(1)-(2) => m=59.5

(1)+(2) => m=sqrt(409)