1. ## even more a²+b²=2c²

Back again

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

$\displaystyle a^2+b^2=2c^2$

we can write a, b & c like this:
$\displaystyle a=m^2+2m-1\ \ \ \ (1)$
$\displaystyle b=m^2-2m-1\ \ \ \ (2)$
$\displaystyle c=m^2+1\ \ \ \ \ \ \ \ \ \ \ \ (3)$
Proof:

$\displaystyle a^2=m^4+4m^3+2m^2-4m+1$
$\displaystyle b^2=m^4-4m^3+2m^2+4m+1$
$\displaystyle c^2=m^4+2m^2+1$

$\displaystyle so\ \ \ a^2+b^2=2m^4+4m^2+2\ \ \ =2c^2\ \ \ Q.E.D.$
$\displaystyle Now\ \ \ 194^2+142^2=2(170^2)$

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

using (1)-(2) we have $\displaystyle \ \ a-b=4m\ \ \4m=194-142=52\ \ \ \ so\ m=13$

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

$\displaystyle (1)\ \ \ a=m^2+2m-1\ \ \ 169+26-1=194$
$\displaystyle (2)\ \ \ b=m^2-2m-1\ \ \ 169-26-1=142$
$\displaystyle (3)\ \ \ c=m^2+1\ \ \ \ \ \ \ \ \ \ \ 169+1=170$

All good so far but...

$\displaystyle 527^2+289^2=2(425^2)$

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

using (1)-(2) we have $\displaystyle \ \ a-b=4m\ \ \4m=527-289=238\ \ \ \ so\ m=59.5$

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

$\displaystyle (1)\ \ \ a=m^2+2m-1\ \ \ 3540.25+119-1=3658.25\ \ \ \ not\ the\ expected\ 527$
$\displaystyle (2)\ \ \ b=m^2-2m-1\ \ \ 3540.25-119-1=3420.25\ \ \ \ not\ the\ expected\ 289$
$\displaystyle (3)\ \ \ c=m^2+1\ \ \ \ \ \ \ \ \ 3540.25+1=3541.25\ \ \ \ not\ the\ expected\ 425$

What am I doing wrong or misunderstanding?

Many thanks

Pro

2. ## Re: even more a²+b²=2c²

Ignore this post.

Pro,

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

3. ## Re: even more a²+b²=2c²

Originally Posted by SammyS
Ignore this post.

ah! just noticed your reason for editing

No problem SammyS, lol

Thanks

Pro

4. ## Re: even more a²+b²=2c²

Originally Posted by procyon
Back again

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

$\displaystyle a^2+b^2=2c^2$

we can write a, b & c like this:
$\displaystyle a=m^2+2m-1\ \ \ \ (1)$
$\displaystyle b=m^2-2m-1\ \ \ \ (2)$
$\displaystyle c=m^2+1\ \ \ \ \ \ \ \ \ \ \ \ (3)$
Proof:

$\displaystyle a^2=m^4+4m^3+2m^2-4m+1$
$\displaystyle b^2=m^4-4m^3+2m^2+4m+1$
$\displaystyle c^2=m^4+2m^2+1$

$\displaystyle so\ \ \ a^2+b^2=2m^4+4m^2+2\ \ \ =2c^2\ \ \ Q.E.D.$
$\displaystyle Now\ \ \ 194^2+142^2=2(170^2)$

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

using (1)-(2) we have $\displaystyle \ \ a-b=4m\ \ \4m=194-142=52\ \ \ \ so\ m=13$

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

$\displaystyle (1)\ \ \ a=m^2+2m-1\ \ \ 169+26-1=194$
$\displaystyle (2)\ \ \ b=m^2-2m-1\ \ \ 169-26-1=142$
$\displaystyle (3)\ \ \ c=m^2+1\ \ \ \ \ \ \ \ \ \ \ 169+1=170$

All good so far but...

$\displaystyle 527^2+289^2=2(425^2)$

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

using (1)-(2) we have $\displaystyle \ \ a-b=4m\ \ \4m=527-289=238\ \ \ \ so\ m=59.5$

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

$\displaystyle (1)\ \ \ a=m^2+2m-1\ \ \ 3540.25+119-1=3658.25\ \ \ \ not\ the\ expected\ 527$
$\displaystyle (2)\ \ \ b=m^2-2m-1\ \ \ 3540.25-119-1=3420.25\ \ \ \ not\ the\ expected\ 289$
$\displaystyle (3)\ \ \ c=m^2+1\ \ \ \ \ \ \ \ \ 3540.25+1=3541.25\ \ \ \ not\ the\ expected\ 425$

What am I doing wrong or misunderstanding?

Many thanks

Pro
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 $\displaystyle m=1+\sqrt{291}$, so $\displaystyle a=293+4\sqrt{291}\,,$ and $\displaystyle c=293+2\sqrt{291}\,,$

... etc.

You can pick an integer, m, then get a, b, and c, so that $\displaystyle a^2+b^2=2c^2\,,$ but simply finding a, b, c that satisfy $\displaystyle a^2+b^2=2c^2$ does not ensure that there is a number m that generates a, b, and c.

5. ## [solved] Re: even more a²+b²=2c²

You can pick an integer, m, then get a, b, and c, so that $\displaystyle \ \ a^2+b^2=2c^2$ but simply finding a, b, c that satisfy $\displaystyle \ \ a^2+b^2=2c^2$ does not ensure that there is a number m that generates a, b, and c.
I've come to the same conclusion but still can't really get my head around it
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)