How can I find polynomials a,b and c that satisfy a^2+b^2=c^2+1?

>>8624814

Especially given that it's supposed to be the same woman's hand, that hand is freakishly angular and large. Women have these little thin stubby hands, most of the time.

Initially I found your language confusing, believing that you were confusing polynomials themselves with numbers, and I was about to scold you about it, but your expansion sounds like you just want to plug in and square polynomials in your thing, and find a general family of solutions.

Polynomials can be tricky. There are such things as sparse polynomials, so for all I know, your thing is hard as well. Anyway I can't be fussed anymore, I was just caught by that woman's freakishly large alien-hand.

a=2t, b=2t^2-1, c=2t^2 would be one family of solutions.

In this case I just looked at two consecutive squares (b^2 and c^2 in this case), and pondered what requirements are necessary for their difference to be one less than a square (a).

If you move to b and c having a difference of 2, you get:
a=2t+1, b=t^2+t-1, c=t^2+t+1

File: OrthoPythag.png (111KB, 986x527px) Image search: [Google]

111KB, 986x527px

>>8625118
>>8625238

Thank you for your replies

a=2t, b=2t^2-1, c=2t^2

This is very nice, thank you! I wonder if this is the only such solution up to some fiddling. You could apply a generalized "orthogonal" transform, for example. See pic from
"Pythagorean Descent" by Keith Conrad.

It's frustrating poking around for help on the web. I find a reference that seems like it might be helpful, but they always seem to jump to advanced math topics which seem detached from any practical application.

[math]a=2st+s+1\\b=2st^2+2st+2t+1\\c=2st^2+2st+2t+s+1[/math]

>>8625768
Excellent!
I got something similar generalizing
>>8625238
Assuming b and c differ by a constant. Any feeling about whether more generality is possible?

I was thinking about finding solutions of higher degree and wonder if this would work...

Suppose we have an initial triple of polynomials
polynomials with
a^2+b^2=c^2+1,
and we look for another of the form (a+S*A,b+S*B,c+S*C),
for polynomials S,A,B and C to be determined. The condition
(a+S*A)^2+(b+S*B)^2-(c+S*C)^2 = a^2+b^2-c^2
can be solved formally for S, yielding
S = -(2 (a A + b B - c C))/(A^2 + B^2 - C^2).
Since S needs to be a polynomial, we need A^2+B^2-C^2=k (a constant).
And we know how to do that for 2nd degree A B and C from this thread. The new triple is
a+s*A =a-2*A*(a A + b B - c C)/k
b+s*B =b-2*B*(a A + b B - c C)/k
c+s*C =c-2*C*(a A + b B - c C)/k

If A,B and C are degree 2, then the new
triple will be of degree 4 more than the old.

File: Screenshot_20170124-195744.png (2MB, 1440x2560px) Image search: [Google]

2MB, 1440x2560px

>>8628215
Could you make a topology of pic with that?

>>8624814

So I finally figured it out, after resorting to brute force algebra. The following code produces quadratic polynomials a,b and c such that
a^2+b^2-c^2 = 1. In the end, there are 4 free parameters. You should be able to get all of them this way. I can't say if this is any different from a combination of this
>>8625768
and this
>>8625677
Since this combination would also have 4 degrees of freedom. Thanks for your help /sci/.

clear all

h=randn(1);
t=randn(1);
p=randn(1);
q=randn(1);

d = p*t+q*t+1;
e = p^2-q^2+1;
r = -h^2*e*(p-q-t)/d ;
A = 2*h*(p*(t^2+1)+q*(t^2-1))/d;
B = h*((-p^2+q^2+1)*(t^2+1)+4*q*t)/d;
C = h*((p^2-q^2-1)*(t^2-1)+4*p*t)/d;

% 3 coefficeints for each
% of polynomials a,b and c, high order to low order
a = [2*r*t A (p^2-q^2-1)/e];
b = [r*(1-t^2) B 2*p/e];
c = [r*(1+t^2) C 2*q/e];

f = conv(a,a)+conv(b,b)-conv(c,c);

max(abs( f-[0 0 0 0 1] ))