How to create a private key using $n,e,d,p,q$ where $n$ is prime

Question

Someone has incorrectly created a public/private key pair where $N$ (modulus) is a prime number.

This makes $p$ and $q$ trivial as they they are $n$ and $1$.

$d$ is solved from $d \cdot e = \bmod (p-1)(q-1)$ which becomes $d \cdot e = \bmod (n-1)$

I want to create the private key knowing this information but no PEM generators I've found correctly handle $p$ or $q$ being $1$ because when you need to find $e_1$ or $e_2$ you are looking for $d \bmod (p-1)$ and $(p-1)$ [or $q-1$] would be zero and it crashes.

I've tried python's RSA.construct and attempted to make a asn1.cnf but I am unable to calculated $e_1/e_2$.

I am confident it can be cracked though because RSA decrypt is just ciphertext to the $d$-power, $\bmod n$.

Answer 1

You already stated the answer yourself: You know $N$ (which is prime by your assumption) and from the public key $e$ you can compute $d = e^{-1} \bmod N-1$, e.g. using the EEA.

With this $d$ you can then decrypt any ciphertext $c$ into the corresponding $m = c^d \bmod N$.

Answer 2

FWIW, (desktop=Oracle) Java accepts and successfully uses a n,d-only/non-CRT-form RSA private key with an invalidly prime n, and it even generates and reads back what it claims to be a PKCS8/PKCS1 encoding (that could easily be PEMified) -- by setting the unused fields to zero! In my mind that takes Postelianism too far by half. But this does 'work', in the very limited sense of recovering the correct plaintext:

    BigInteger p = new BigInteger (1024, 128, new SecureRandom());
    BigInteger e = BigInteger.valueOf(3), d; 
    while( !p.subtract(BigInteger.ONE).gcd(e).equals(BigInteger.ONE) ) e.add(BigInteger.valueOf(2));
    d = e.modInverse(p.subtract(BigInteger.ONE));

    KeyFactory fact = KeyFactory.getInstance("RSA");
    PublicKey pub = fact.generatePublic(new RSAPublicKeySpec(p,e));
    System.out.println (DatatypeConverter.printHexBinary(pub.getEncoded()));
    PrivateKey prv = fact.generatePrivate(new RSAPrivateKeySpec(p,d));
    System.out.println (DatatypeConverter.printHexBinary(prv.getEncoded()));
    PrivateKey pr2 = fact.generatePrivate(new PKCS8EncodedKeySpec(prv.getEncoded()));

    Cipher ciph = Cipher.getInstance("RSA");
    ciph.init(Cipher.ENCRYPT_MODE, pub);
    byte[] enc = ciph.doFinal("TEST".getBytes()); // FOR TEST charset 
    ciph.init(Cipher.DECRYPT_MODE, pr2);
    System.out.println (new String (ciph.doFinal(enc))); // doesn't matter 

-->
30819D300D06092A864886F70D010101050003818B0030818702818100C906C38BD8F2790F34D7DD453F8D8F26309E13748F6C394EF9F3698E35089D0447987DAF31AA207203358F27E9DF009BFBAA530386C30188076A9A085FA81E48087EB9817DDBBBFE7878B36DF7C6B74DCAC16E5492F2DC2F346E6C2E03585D59CDF2F918EBED73CA8F6B5C449F68A95EECB5B0798B89254F72E077BAAEF4A583020103
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
TEST

It is course totally insecure, as already described.