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