RSA简单实现公钥私钥正确为何加密解密后不一致？

Question

import java.math.BigInteger;
import java.security.SecureRandom;
import java.util.Date;

public class RSA {
private BigInteger p,q,n,d;
private BigInteger e=new BigInteger("65537");
private static BigInteger y;
private static BigInteger x;

    private void getPrime(int length){
        SecureRandom random=new SecureRandom();
        random.setSeed(new Date().getTime());
        while((p=BigInteger.probablePrime(length, random)).isProbablePrime(1)==false)
        {
            continue;
        }
        while((q=BigInteger.probablePrime(length, random)).isProbablePrime(1)==false)
        {
            continue;
        }
    }

    public static BigInteger e_gcd(BigInteger a,BigInteger b){
        if(b.intValue()==0){
            x=new BigInteger("1");
            y=new BigInteger("0");
            return a;
        }
        BigInteger ans=e_gcd(b, a.mod(b));
        BigInteger tmp=x;
        x=y;
        y=tmp.subtract(a.divide(b).multiply(y));

// System.out.println("tmp= "+tmp);
// System.out.println("gcd a="+a+" gcd b="+b+" gcd x="+x+" gcd y="+y);
// //System.out.println("gcd y="+y);
//

return ans;
}
BigInteger cald(BigInteger e,BigInteger n){
BigInteger gcd=e_gcd(e, n);
if(BigInteger.ONE.mod(gcd).intValue()!=0)
return new BigInteger("-1");
x.multiply(BigInteger.ONE.divide(gcd));
n=n.abs();
BigInteger ans=x.mod(n);
if(ans.compareTo(BigInteger.ZERO)<0)
ans.add(n);
return ans;
}

    public BigInteger getp(){
        return p;
    }
    public BigInteger getq(){
        return q;
    }
    public BigInteger getn(){
        return n;
    }
    public BigInteger gete(){
        return e;
    }
    public BigInteger getd(){
        return d;
    }

    public RSA(int length){
        getPrime(length/2);
        n=p.multiply(q);
        d=cald(e, n);
        //System.out.println("test : "+e.multiply(d).mod(n));
    }
    public RSA() {
        getPrime(128);
        n=p.multiply(q);
        //d=cald(e, n);
        BigInteger tmp=p.subtract(BigInteger.ONE).multiply(q.subtract(BigInteger.ONE));

        d=e.modPow(tmp.subtract(BigInteger.ONE),n );
    }
public static void main(String[] args) {
    // TODO Auto-generated method stub

    RSA rsa=new RSA(20);
    System.out.println("P= " +rsa.getp());
    System.out.println("Q= " +rsa.getq());
    System.out.println("E= " +rsa.gete());
    System.out.println("N= " +rsa.getn());
    System.out.println("D= " +rsa.getd());
    BigInteger test=new BigInteger("100");
    System.out.println("test:"+rsa.gete().multiply(rsa.getd()).mod(rsa.getn()));
    BigInteger testc=test.modPow(rsa.getd(),rsa.getn());
    testc=testc.modPow(rsa.gete(),rsa.getn());
    boolean flag =testc.equals(test);
    System.out.println(flag);

}

}

代码如上，e*d=1(mod n)成立了，为何在加解密之后不一致呢？？

Answer 1

https://zhidao.baidu.com/question/753614306264418764.html