On 2013-05-21, Martin Brown wrote:
On 21/05/2013 09:11, Jasen Betts wrote:
On 2013-05-20, Martin Brown wrote:
On 20/05/2013 10:49, Jasen Betts wrote:
On 2013-05-20, Martin Brown wrote:
On 19/05/2013 22:09, Tom Del Rosso wrote:

The algebraic solution is in the form

A = F(a,b)
B = G(a,b)

Where A,B,a,b are all integers with some of them non-zero.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
In the general case they are all non-zero.
Your task is to find the functions F(), G().

functions?

F(a,b) = sqrt(a^2+b^2)
G=F

that seems way too easy

sqrt(a^2+b^2) isn't an integer except in a handful of special cases.

hey! stop changing the rules!

I didn't change the rules. Though I left the question open at first.

you said sqrt(a^2+b^2) AKA "c" was an integer in your first
post:

Martin Brown wrote:

given a^2 + b^2 = c^2 where a,b,c are integers


What was c^2 for? is it an attempt at misdirection?
the problem with that is it makes finding a valid answer easier.

I am actually not going to post the answer here and would ask that
anyone who does work it out do likewise. It is obvious once you see it.

the pattern is obviuos, I had to derive 2a^2 + 2b^2 before I believed it.
I posted it before I read this.


--
š‚šƒ 100% natural

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