(CN) RSA Public-Key Encryption and Signature Lab

Outis Yang
Feb 8, 2023 3 min read

SEED Project官网

RSA Public-Key Encryption and Signature Lab

Task 1 Deriving the Private Key

#include <stdio.h>
#include <openssl/bn.h>
#define NBITS 128

void printBN(char *msg, BIGNUM *a){
    char *number_str = BN_bn2hex(a);
    printf("%s %s\n", msg, number_str);
    OPENSSL_free(number_str);
}

int main(){
    //init
    BN_CTX *ctx = BN_CTX_new();
    BIGNUM *p = BN_new();
    BIGNUM *q = BN_new();
    BIGNUM *fai_n = BN_new();
    BIGNUM *n = BN_new();
    BIGNUM *e = BN_new();
    BIGNUM *d = BN_new();
    BIGNUM *res = BN_new();
    BIGNUM *p_1 = BN_new();
    BIGNUM *q_1 = BN_new();

    //get value
    BN_hex2bn(&p, "F7E75FDC469067FFDC4E847C51F452DF");
    BN_hex2bn(&q, "E85CED54AF57E53E092113E62F436F4F");
    BN_hex2bn(&e, "0D88C3");

    //calculate
    BN_sub(p_1, p, BN_value_one());
    BN_sub(q_1, q, BN_value_one());
    BN_mul(n, p, q, ctx);
    BN_mul(fai_n, p_1, q_1, ctx);

    // check if e and fai(n) is relatively prime
    BN_gcd(res, fai_n, e, ctx);
    if (!BN_is_one(res)){
        printf("Error: e and fai(n) is not relatively prime \n ");
        exit(0);
    }

    BN_mod_inverse(d, e, fai_n, ctx);

    printBN("public key e=\t", e);
    printBN("public key n=\t", n);
    printBN("private key d=\t", d);
    return 0;
}
image-20230208105705992

Task 2 Encrypting a Message

#include <stdio.h>
#include <openssl/bn.h>
#define NBITS 128

void printBN(char *msg, BIGNUM *a){
    char *number_str = BN_bn2hex(a);
    printf("%s %s\n", msg, number_str);
    OPENSSL_free(number_str);
}

int main(){
    //init
    BN_CTX *ctx = BN_CTX_new();
    BIGNUM *n = BN_new();
    BIGNUM *e = BN_new();
    BIGNUM *d = BN_new();
    BIGNUM *msg = BN_new(); //massage
    BIGNUM *p = BN_new();   //plaintxt
    BIGNUM *c = BN_new();   //cyphertxt

    //get value
    BN_hex2bn(&n, "DCBFFE3E51F62E09CE7032E2677A78946A849DC4CDDE3A4D0CB81629242FB1A5");
    BN_hex2bn(&e, "010001");
    BN_hex2bn(&d, "74D806F9F3A62BAE331FFE3F0A68AFE35B3D2E4794148AACBC26AA381CD7D30D");
    BN_hex2bn(&msg, "4120746f702073656372657421");

    //calculate
    BN_mod_exp(c, msg, e, n, ctx);
    BN_mod_exp(p, c, d, n, ctx);
    
    printBN("cyphertxt:", c);
    printBN("plaintxt:", p);
    return 0;
}
image-20230208105643641

Task 3 Decrypting a Message

#include<stdio.h>
#include<openssl/bn.h>
#define NBITS 128

void printBN(char *msg, BIGNUM *a){
    char *number_str = BN_bn2hex(a);
    printf("%s %s\n", msg, number_str);
    OPENSSL_free(number_str);
}

int main(){
    //init
    BN_CTX *ctx = BN_CTX_new();
    BIGNUM *n = BN_new();
    BIGNUM *e = BN_new();
    BIGNUM *d = BN_new();
    BIGNUM *p = BN_new(); //plaintxt
    BIGNUM *c = BN_new(); //cyphertxt

    //get value
    BN_hex2bn(&n, "DCBFFE3E51F62E09CE7032E2677A78946A849DC4CDDE3A4D0CB81629242FB1A5");
    BN_hex2bn(&e, "010001");
    BN_hex2bn(&d, "74D806F9F3A62BAE331FFE3F0A68AFE35B3D2E4794148AACBC26AA381CD7D30D");
    BN_hex2bn(&c, "8C0F971DF2F3672B28811407E2DABBE1DA0FEBBBDFC7DCB67396567EA1E2493F");

    //calculate
    BN_mod_exp(p, c, d, n, ctx);
    
    printBN("plaintxt:", p);
    return 0;
}
image-20230208105608714

Task 4 Signing a Message

image-20230208111641559 
#include <stdio.h>
#include <openssl/bn.h>
#define NBITS 128

void printBN(char *msg, BIGNUM *a){
    char *number_str = BN_bn2hex(a);
    printf("%s %s\n", msg, number_str);
    OPENSSL_free(number_str);
}

int main(){
    //init
    BN_CTX *ctx = BN_CTX_new();
    BIGNUM *n = BN_new();
    BIGNUM *e = BN_new();
    BIGNUM *d = BN_new();
    BIGNUM *m1 = BN_new();
    BIGNUM *m2 = BN_new();
    BIGNUM *sig1 = BN_new();
    BIGNUM *sig2 = BN_new();

    //get value
    BN_hex2bn(&n, "DCBFFE3E51F62E09CE7032E2677A78946A849DC4CDDE3A4D0CB81629242FB1A5");
    BN_hex2bn(&e, "010001");
    BN_hex2bn(&d, "74D806F9F3A62BAE331FFE3F0A68AFE35B3D2E4794148AACBC26AA381CD7D30D");
    BN_hex2bn(&m1, "49206f776520796f75202432303030");
    BN_hex2bn(&m2, "49206f776520796f75202433303030");

    //calculate
    BN_mod_exp(sig1, m1, d, n, ctx);
    BN_mod_exp(sig2, m2, d, n, ctx);
    
    printBN("signature of m1:", sig1);
    printBN("signature of m2:", sig2);
    return 0;
}
image-20230208111947209

可以发现，虽然信息只做了轻微的改动，签名结果却发生很大的变化。

Task 5 Verifying a Signature

image-20230208113201340 
#include <stdio.h>
#include <openssl/bn.h>
#define NBITS 128

void printBN(char *msg, BIGNUM *a){
    char *number_str = BN_bn2hex(a);
    printf("%s %s\n", msg, number_str);
    OPENSSL_free(number_str);
}

int main(){
    //init
    BN_CTX *ctx = BN_CTX_new();
    BIGNUM *n = BN_new();
    BIGNUM *e = BN_new();
    BIGNUM *M = BN_new();
    BIGNUM *m1 = BN_new();
    BIGNUM *m2 = BN_new();
    BIGNUM *sig1 = BN_new();
    BIGNUM *sig2 = BN_new();

    //get value
    BN_hex2bn(&M, "4c61756e63682061206d697373696c652e");
    BN_hex2bn(&n, "AE1CD4DC432798D933779FBD46C6E1247F0CF1233595113AA51B450F18116115");
    BN_hex2bn(&e, "010001");
    BN_hex2bn(&sig1, "643D6F34902D9C7EC90CB0B2BCA36C47FA37165C0005CAB026C0542CBDB6802F");
    BN_hex2bn(&sig2, "643D6F34902D9C7EC90CB0B2BCA36C47FA37165C0005CAB026C0542CBDB6803F");

    //calculate
    BN_mod_exp(m1, sig1, e, n, ctx);
    BN_mod_exp(m2, sig2, e, n, ctx);

    //check if the result is valid
    printf("verifying of signature1:");
    if (BN_cmp(m1, M) == 0)
        printf("valid!\n");
    else
        printf("invalid!\n");
    
    printf("verifying of signature2:");
    if (BN_cmp(m2, M) == 0)
        printf("valid!\n");
    else
        printf("invalid!\n");

    printBN("m1:", m1);
    printBN("m2:", m2);
    printBN("M:", M);
    return 0;
}
image-20230208115532253

当签名没有损坏时，可以验证成功。即使签名只变动了一位，也会使得验证失败。

Seedlab
Outis Yang
Outis Yang
2024 Undergraduate in Cyberspace Security

My research interests include Internet of Vehicles(IoV), Penetration Testing and Security research.