Time for T with U (Posted on 2014-01-19)

	Submitted by broll
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I won't bore with theory, just work through the equivalences. We start with: P1=1 P2=4x+3 P3=16x^2+20x+5 P4=64x^3+112x^2+56x+7 ... Q1=1 Q2=4x+1 Q3=16x2+12x+1 Q4=64x3+80x2+24x+1 ... First, conduct the squaring operation to confirm the relation: (4n+3)^2*n+1 = (n+1) (4n+1)^2 (16n^2+20n+5)^2*n+1 = (n+1) (16n^2+12n+1)^2 (64n^3+112n^2+56n+7)^2*n+1 = (n+1) (64n^3+80n^2+24n+1)^2 (256n^4+576n^3+432n^2+120n+9)^2*n+1 = (n+1) (256n^4+448n^3+240n^2+40n+1)^2 ...etc. Note that all these equations are of the form k*a^2=(k-1)b^2+1, with n=k, a crucial point. Second, cross-relate to the Chebyshev polynomials (just the P series to show the method) ChebyshevTn(x): T0(x)=1 T1(x)= x T2(x)= 2x^2-1 T3(x)= 4x^3-3x T4(x)= 8x^4-8x^2+1 ChebyshevUn(x) U0(x)= 1 U1(x)= 2x U2(x)= 4x^2-1 U3(x)= 8x^3-4x U4(x)= 16x^4-12x^2+1 Given that P2=4x+3; let s be the smallest solution, greater than 1, that solves the corresponding Pell-like equation, with s=T1[x]=(2x+1)=U1[x], k=(x+1), m=2; where [x] denotes the base value before multiplication of T1(x) and U1(x); k(k-1) = D in the canonical equation a^2=Db^2+1; and m is the auxiliary multiplier corresponding to D (as it happens, m = 2 in such cases). (4x+3) = (x+1)*2+(2x+1) 16x^2+20x+5 = (x+1)*((2x+1)*2)*2+2*(2x+1)^2-1 64x^3+112x^2+56x+7 = (x+1)* (4*(2x+1)^2-1)*2+4*(2x+1) ^3-3*(2x+1) 256x^4+576x^3+432x^2+120x+9 = (x+1)*(8*(2x+1)^3-4*(2x+1))*2+8*(2x+1)^4-8*(2x+1)^2+1 1024x^5+2816x^4+2816x^3+1232x^2+220x+11 = (x+1)*2*(16*(2x+1)^4-12*(2x+1)^2+1)+ 16*(2x+1)^5-20*(2x+1)^3+5*(2x+1) etc. In general: Pn[s] = 2*Un[s](x+1)+Tn[s], and we are done.