我试图找到Java(本机或BouncyCastle提供程序)实现,使用给定的参数{e,n,d}在PKCS#1中生成RSA私钥。
Dan Boneh有paper描述了这样做的算法。该解决方案在PyCrypto(Python)中可用,并且Mounir IDRASSI发布了一个独立的utility,它在SFM格式(n,e,d)和CRT格式(p,q,dp)之间转换RSA密钥,dq,u),反之亦然。但是我无法找到任何可以用于Java的东西。
更新:我在https://github.com/martinpaljak/RSAKeyConverter/blob/master/src/opensc/RSAKeyConverter.java
找到了此类实施答案 0 :(得分:1)
我在this回答中提供了一些代码,我将在此处重现:
/**
* Find a factor of n by following the algorithm outlined in Handbook of Applied Cryptography, section
* 8.2.2(i). See http://cacr.uwaterloo.ca/hac/about/chap8.pdf.
*
*/
private static BigInteger findFactor(BigInteger e, BigInteger d, BigInteger n) {
BigInteger edMinus1 = e.multiply(d).subtract(BigInteger.ONE);
int s = edMinus1.getLowestSetBit();
BigInteger t = edMinus1.shiftRight(s);
for (int aInt = 2; true; aInt++) {
BigInteger aPow = BigInteger.valueOf(aInt).modPow(t, n);
for (int i = 1; i <= s; i++) {
if (aPow.equals(BigInteger.ONE)) {
break;
}
if (aPow.equals(n.subtract(BigInteger.ONE))) {
break;
}
BigInteger aPowSquared = aPow.multiply(aPow).mod(n);
if (aPowSquared.equals(BigInteger.ONE)) {
return aPow.subtract(BigInteger.ONE).gcd(n);
}
aPow = aPowSquared;
}
}
}
public static RSAPrivateCrtKey createCrtKey(RSAPublicKey rsaPub, RSAPrivateKey rsaPriv) throws NoSuchAlgorithmException, InvalidKeySpecException {
BigInteger e = rsaPub.getPublicExponent();
BigInteger d = rsaPriv.getPrivateExponent();
BigInteger n = rsaPub.getModulus();
BigInteger p = findFactor(e, d, n);
BigInteger q = n.divide(p);
if (p.compareTo(q) > 1) {
BigInteger t = p;
p = q;
q = t;
}
BigInteger exp1 = d.mod(p.subtract(BigInteger.ONE));
BigInteger exp2 = d.mod(q.subtract(BigInteger.ONE));
BigInteger coeff = q.modInverse(p);
RSAPrivateCrtKeySpec keySpec = new RSAPrivateCrtKeySpec(n, e, d, p, q, exp1, exp2, coeff);
KeyFactory kf = KeyFactory.getInstance("RSA");
return (RSAPrivateCrtKey) kf.generatePrivate(keySpec);
}