• The Journal of the Institute of Internet, Broadcasting and Communication
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• v.12 no.5
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• pp.185-189
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• 2012

It is almost impossible to directly find the prime factor, p,q of a large semiprime, n=pq. So Most of the integer factorization algorithms uses a indirect method that find the prime factor of the p=GCD(a-b,n),q=GCD(a+b,n) after getting the congruence of squares of the $a^2{\equiv}b^2$(mod n). Many methods of getting the congruence of squares have proposed, but it is not easy to get with RSA number of greater than a 100-digit number. This paper proposes a fast algorithm to get the congruence of squares. The proposed algorithm succeeded in getting the congruence of squares to a 19-digit number.

• The Journal of the Institute of Internet, Broadcasting and Communication
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• v.13 no.6
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• pp.221-228
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• 2013

It is impossible directly to find a prime number p,q of a large semiprime n = pq using Trial Division method. So the most of the factorization algorithms use the indirection method which finds a prime number of p = GCD(a-b, n), q=GCD(a+b, n); get with a congruence of squares of $a^2{\equiv}b^2$ (mod n). It is just known the fact which the area that selects p and q about n=pq is between $10{\cdots}00$ < p < $\sqrt{n}$ and $\sqrt{n}$ < q < $99{\cdots}9$ based on $\sqrt{n}$ in the range, [$10{\cdots}01$, $99{\cdots}9$] of $l(p)=l(q)=l(\sqrt{n})=0.5l(n)$. This paper proposes the method that reduces the range of p using information obtained from n. The proposed method uses the method that sets to $p_{min}=n_{LR}$, $q_{min}=n_{RL}$; divide into $n=n_{LR}+n_{RL}$, $l(n_{LR})=l(n_{RL})=l(\sqrt{n})$. The proposed method is more effective from minimum 17.79% to maxmimum 90.17% than the method that reduces using $\sqrt{n}$ information.

• The Journal of the Institute of Internet, Broadcasting and Communication
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• v.11 no.2
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• pp.107-112
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• 2011

It is very difficult to factorize composite number, $n=pq$ to integer factorization, p and q that is almost similar length of digits. Integer factorization algorithms, for the most part, find ($a,b$) that is congruence of squares ($a^2{\equiv}b^2$ (mod $n$)) with using factoring(factor base, B) and get the result, $p=GCD(a-b,n)$, $q=GCD(a+b,n)$ with taking the greatest common divisor of Euclid based on the formula $a^2-b^2=(a-b)(a+b)$. The efficiency of these algorithms hangs on finding ($a,b$) and deciding factor base, B. This paper proposes a efficient algorithm. The proposed algorithm extracts B from integer factorization with 3 digits prime numbers of $n+1$ and decides f, the combination of B. And then it obtains $x$(this is, $a=fxy$, $\sqrt{n}$ < $a$ < $\sqrt{2n}$) from integer factorization of $n-2$ and gets $y=\frac{a}{fx}$, $y_1$={1,3,7,9}. Our algorithm is much more effective in comparison with the conventional Fermat algorithm that sequentially finds $\sqrt{n}$ < $a$.

• The Journal of the Institute of Internet, Broadcasting and Communication
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• v.11 no.4
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• pp.157-164
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• 2011

It is very difficult problem to factorize composite number. Integer factorization algorithms, for the most part, find ($a,b$) that is congruence of squares ($a^2{\equiv}b^2$(mode $n$)) with using factoring(factor base, B) and get the result, $p=GCD(a-b,n)$, $q=GCD(a+b,n)$ with taking the greatest common divisor of Euclid based on the formula $a^2-b^2=(a-b)(a+b)$. The efficiency of these algorithms hangs on finding ($a,b$). Fermat's algorithm that is base of congruence of squares finds $a^2-b^2=n$. This paper proposes the method to find $a^2-b^2=kn$, ($k=1,2,{\cdots}$). It is supposed $b_1$=0 or 5 to be surely, and b is a double number. First, the proposed method decides $k$ by getting kn that satisfies $b_1=0$ and $b_1=5$ about $n_2n_1$. Second, it decides $a_2a_1$ that satisfies $a^2-b^2=kn$. Third, it figures out ($a,b$) from $a^2-b^2=kn$ about $a_2a_1$ as deciding $\sqrt{kn}$ < $a$ < $\sqrt{(k+1)n}$ that is in $kn$ < $a^2$ < $(k+1)n$. The proposed algorithm is much more effective in comparison with the conventional Fermat algorithm.

• Journal of the Korea Society of Computer and Information
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• v.19 no.7
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• pp.71-76
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• 2014

There is to be virtually impossible to solve the very large digits of prime number p and q from composite number n=pq using integer factorization in typical public-key cryptosystems, RSA. When the public key e and the composite number n are known but the private key d remains unknown in an asymmetric-key RSA, message decryption is carried out by first obtaining ${\phi}(n)=(p-1)(q-1)=n+1-(p+q)$ and then using a reverse function of $d=e^{-1}(mod{\phi}(n))$. Integer factorization from n to p,q is most widely used to produce ${\phi}(n)$, which has been regarded as mathematically hard. Among various integer factorization methods, the most popularly used is the congruence of squares of $a^2{\equiv}b^2(mod\;n)$, a=(p+q)/2,b=(q-p)/2 which is more commonly used then n/p=q trial division. Despite the availability of a number of congruence of scares methods, however, many of the RSA numbers remain unfactorable. This paper thus proposes an algorithm that directly and immediately obtains ${\phi}(n)$. The proposed algorithm computes $2^k{\beta}_j{\equiv}2^i(mod\;n)$, $0{\leq}i{\leq}{\gamma}-1$, $k=1,2,{\ldots}$ or $2^k{\beta}_j=2{\beta}_j$ for $2^j{\equiv}{\beta}_j(mod\;n)$, $2^{{\gamma}-1}$ < n < $2^{\gamma}$, $j={\gamma}-1,{\gamma},{\gamma}+1$ to obtain the solution. It has been found to be capable of finding an arbitrarily located ${\phi}(n)$ in a range of $n-10{\lfloor}{\sqrt{n}}{\rfloor}$ < ${\phi}(n){\leq}n-2{\lfloor}{\sqrt{n}}{\rfloor}$ much more efficiently than conventional algorithms.