Let E all triplets as X=(p,q,r) such as p^2+q^2=r^2. We define the application f of E dans C complex as X in E f(X)=(p+Iq)/r=Z. Calculate abs(Z). Show that in E the law noted * defined by
X1*X2=(p1*p2-q1*q2,p2*q1+p1*q2,r1*r2) is an internal law. Calculate f(X1*X2). Then if X0=(3,4,5), find
X0*X0, X0*(X0*X0).Thank you for the help.

how to show which region of the plan belongs to the argument points between 0 and Pi/2 and the module points between 0 and 2 ?

Any integer in [129,129+13^2[ 1s written as the sum of squares of separate integers equal to or less than 12.
examples:: 129=10^2+5^2+2^2, 130=11^2+3^2... 132=9^2+5^2+4^2+3^2+1^2...
179=12^2+5^2+3^2+1^2...297=12^2+10^2+7^2+2^2. Thank you.

restart; F := rsolve({16*s(n+1) = 2+12*s(n)-2*s(n-1), s(1) = 1, s(2) = 5/8}, s); Error, (in s) cannot determine if this expression is true or false: n

f := proc (a, b) options operator, arrow; (1/2)*b+(1/2)*arccos(sin(2*a-b)/tan(b)) end proc; dis := proc (A, B) options operator, arrow; sqrt(inner(A-B, A-B)) end proc; bisA := proc (A, B, C) local b, c, M; b, c := dis(A, C), dis(A, B); M := (b*B+C*c)/(b+c); x*(A[2]-M[2])+y*(M[1]-A[1])+A[1]*M[2]-A[2]*M[1] end proc; P := proc (a0, b0) local a, b, c, p1, p2, p3, p4, r, II; a := evalf(a0); b := evalf(b0); c := f(a, b); if b0-0.1e-2