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A new algorithm is derived via derived forms and lifting to related discriminants.",{"@graph":14,"@context":72},[15,34,55],{"@type":16,"itemListElement":17},"BreadcrumbList",[18,23,27,31],{"item":19,"name":20,"@type":21,"position":22},"https://docshare.wps.com","Home","ListItem",1,{"item":24,"name":25,"@type":21,"position":26},"https://docshare.wps.com/document/","Document",2,{"item":28,"name":29,"@type":21,"position":30},"https://docshare.wps.com/document/research-report/","Research & 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square-free decomposition of integers using class groups  1 / 16   \nThanks to the organizers  \nFigure: Cool guy  \n Erik Mulder  Fast square-free decomposition of integers using class groups  2 / 16   \nIntroduction  \nDefinition  \nSquare-free decomposition problem: Let n = a2 b with a, b ∈ N and b square-free. Find a and b.  \n Who doesn’t like integer factorization?  \n Cryptographic systems that use n = p2 q or n = pk q rely on the hardness of factoring of n.  \n Computing ring of integers of number field  \n Computing endomorphism ring of an elliptic curve over a finite field  \n Erik Mulder  Fast square-free decomposition of integers using class groups  3 / 16   \nMain result  \n\n| \u003Cbr>Theorem |\n| --- |\n| \u003Cbr>Assume some heuristic assumptions. Fix 0 ≤ α ≤ 1. Then for all integers n = a2 b with b = nα, there is an algorithm that finds the square-free decomposition of n in expected time:\u003Cbr>O (Lb[1/2, 1]) = O (Ln[1/2, √α]) ,\u003Cbr>where\u003Cbr>Lb [α, c] = e (c+o(1)) ln(b)α (ln ln(b))1 − α . |\n\nIf a, b are distinct primes of roughly the same cryptographic size, then this is the current fastest method.  \n Erik Mulder  Fast square-free decomposition of integers using class groups  4 / 16   \nBinary quadratic forms  \n f (x, y) = ax2 + bxy + cy2 = (a, b, c), where a, b, c ∈ Z.  \n D = Df = b2 − 4ac is the discriminant of f. We always have D \u003C 0.  f is primitive if gcd(a, b, c) = 1.  \n f , g are equivalent if there exists A = ( p qr s ) ∈ Γ such that f (x, y) = g(A · (x, y)T ) = g(px + qy , rx + sy) . Then Df = Dg .  \n If |b| \u003C a \u003C c, then f is reduced. Every form is equivalent to a unique reduced form.  \n C (D) is the class group of forms of discriminant D . With composition  \nas its group operation. If D = 0 mod 4, then eD = (1, 0 , −~~4~~D ) .  The class number h (D) ≈ √ D is the order of C (D) .  \n Why not use ideals instead of forms?  \n Erik Mulder  Fast square-free decomposition of integers using class groups  5 / 16   \nSchnorr-Lenstra factorization  \n Let n ∈ N , possibly square-free  Take a random f ∈ C (−4n)  \n Construct k, consisting of powers of all primes up to some B  Pray that fk = e−4n, i.e. h (−4n) is smooth  \n If so, try to construct g of order 2  \n Then g is of the form (d , 0 , ~~n~~d )  \n The complete factorization of n can be found this way  \n If h (−4n) is not smooth, try C (−4ns) instead for some small square-free s  \n Quite similar to the ECM  \n Erik Mulder  Fast square-free decomposition of integers using class groups  6 / 16   \nDerived forms I  \nDefinition  \nLet p be a prime and D a discriminant. We say f ∈ C (Dp2 ) is derived from g ∈ C (D) if there exists a 2 × 2 integer matrix A with det(A) = p such that f (x, y) = g(A · (x, y)T ) .  \nProposition  \n For every form f ∈ C (Dp2 ) there exists a unique g ∈ C (D) up to equivalence such that f is derived from g.  \n For each g = (a, b, c) ∈ C (D) there are exactly p −􀀀Dp􀀁 inequivalent primitive forms in C (Dp2 ) that are derived from g.  \n These forms are:  \n(ap2 , p (b + 2ah), ah2 + bh + c) for 0 ≤ h ≤ p − 1 and (a, bp , cp2 )  \n Erik Mulder  Fast square-free decomposition of integers using class groups  7 / 16   \nDerived forms II  \nCorollary  \nh (Dp2 ) = h (D) · (p − 􀀒Dp􀀓 ) .  \n\n| \u003Cbr>Proposition |\n| --- |\n| \u003Cbr>If f1 , f2 ∈ C (Dp2 ) are derived from g1 , g2 ∈ C (D) respectively, then f1 · f2 is derived from g1 · g2 . |\n\n Erik Mulder  Fast square-free decomposition of integers using class groups  8 / 16   \nThe new square-free decomposition algorithm I  \nFrom now on, n = p2 b, with p prime.  \nProposition  \nSuppose g ∈ C (−4n) is derived from e−4b = (1, 0 , b) ∈ C (−4b) and g ≁ e−4n = (1, 0 , n) . Furthermore, suppose that g is reduced and b > p2. Then  \ng = (p2 , 2pk , k2 + b)  \nfor some −p/2 ≤ k ≤ p/2 .  \nLemma  \n|  |\n| --- |\n| \u003Cbr>Suppose g ∈ C (−4n) is derived from e−4b . Let r be a prime with r  p. Lift g to some h ∈ C (−4nr2 ) . Then l = hr−(−4n |\n\n r ) is n","cbCaimMpQHbtSUfr","https://ap.wps.com/l/cbCaimMpQHbtSUfr","pdf",340866,16,"English","# Introduction\n## Definition\n# Main result\n# Binary quadratic forms\n## Schnorr-Lenstra factorization\n# Derived forms I\n## Derived forms II\n# The new square-free decomposition algorithm I\n# The new square-free decomposition algorithm II","[{\"question\":\"What does square-free decomposition mean for an integer n?\",\"answer\":\"It writes n = a^2 b with a,b ∈ N and b square-free, then asks for a and b.\"},{\"question\":\"What is the main algorithmic contribution under heuristic assumptions?\",\"answer\":\"For n = a^2 b with b = n^α, it provides an expected-time bound O(Lb[1/2,1]) with an equivalent expression O(Ln[1/2,√α]).\"},{\"question\":\"How do derived forms and lifting help in the new decomposition algorithm?\",\"answer\":\"Derived forms relate class groups of discriminants D and Dp^2, and the method lifts elements to a larger discriminant to force smoothness conditions, enabling recovery of a factor p^2 of n.\"}]","Fast square-free decomposition of integers using class groups - Erik Mulder - 17 July 2024 | PDF",1791073121]