Updated index.rst

index.rst

@@ -458,6 +458,8 @@ Integer Factorization and Discrete logarithm problem traditionally have been a m
 	(*) If Even Goldbach conjecture is True, Primes P and Q can be written as two sum of squares one per prime: N = 2n = P + Q = xy = p1 + p2 + p3 + ... + pk = p1a^2 + p1b^2 + p1c^2 + p1d^2 + ... + pkd^2 = SOS(P) + SOS(Q).
 	(*) If Even Goldbach conjecture is True, earlier sum of SOS(P) and SOS(Q) of length 4k could be written as twice of another SOS(R) of length m:
 		N = 2n = P + Q = xy = (length-k random partition of N) p1 + p2 + p3 + ... + pk = (Sum of 4 Squares of each pi of length 4k) p1a^2 + p1b^2 + p1c^2 + p1d^2 + ... + pkd^2 = (length 4k-l) SOS(P) + (length l) SOS(Q) = 2*(r1^2 + r2^2 + r3^2 + ... + rm^2) = (length m) 2*SOS(R).
+	(*) Geometric intuition for SOS formulation of Goldbach conjecture earlier which is a Universal Quadratic Form (https://en.wikipedia.org/wiki/Quadratic_form): Necessary condition for Goldbach conjecture to be true is "4k squares must be coalesceable to 8 squares segregated into 2 equal-area rectangles composed of 4 squares each".
+	(*) Universal Quadratic Form version of Goldbach conjecture earlier leads to 15 and 290 theorems by [Conway–Schneeberger] and [Manjul Bhargava]- https://en.wikipedia.org/wiki/15_and_290_theorems - which stipulate conditions for integer matrix representation of the quadratic form (if an integer matrix of quadratic form represents upto 15 or 290, then it represents all integers) perhaps implying limited number of solutions that could be verified for eariler Quadratic Form algorithmically sufficient to prove Goldbach Conjecture. 
 	(*) In Additive Number Theory, Fermat's Sum of Two Squares Theorem - https://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares - states that Every odd prime p can be written as sum of two squares x^2 and y^2 if p = 1 (mod 4). Such primes are termed Pythagorean Primes. Previous Sum of Squares expansion is a generic case of Fermat's Theorem on Sum of Two Squares.
 	(*) By Fermat's Sum of Two Squares Theorem, Previous partition to Sum of Squares reduction solves a special case of Even Goldbach Conjecture if P = 1 (mod 4), Q = 1 (mod 4) and thus SOS(P) = a1^2 + b1^2 and SOS(Q) = a2^2 + b2^2 => N = 2n = xy = P + Q = SOS(P) + SOS(Q) = a1^2 + b1^2 + a2^2 + b2^2 which is Lagrange Sum of 4 squares.
 (*) Finding factor pair p and q of integer N=pq such that ratio p/q is closest to 1 is non-trivial problem of almost-square factorization of N (factor sides of rectangle of area N are almost equal - an integer equivalent of real square root algorithm). Such an almost-square is best suited for solutions to two ILPs (equated to factors) of square tile packing of rectangle of area N.