Can an RSA private key have several public keys? Ok, so my cryptography lecturer in University posed this question at the end of the RSA key generation lecture as a brain teaser.

I have been thinking about this and I think I have come up with a way ( I am aware it's not practical and very vulnerable - but I would love for you guys to punch holes through it ):

Use RSA to generate public and private key pair.

Make an additive hash function that takes in any input and adds the
ASCII value of each character and gives that as an output. ( I am aware additive hash has horrible distribution, for example, ABC, BAC, ACB all have the same hash output -- but theoretically, this can be used as an advantage to solve our problem )

Now take your public key and generate several possible keys by jumbling up the on and off bits.

Give these jumbled up keys to potential customers.

Put our hash function on the server.

When a customer puts in his jumbled up public-key - Hash function
calculates the hash - which should be similar to original public key generated by RSA. The customer gets access.

Once again I am aware that this is very vulnerable and terrible. It cant be used in the real world. But does it answer the question my lecturer posed?

EDIT:

I am sorry for the confusion. This question has nothing to do with collisions and exploiting the maths ( Those are definitely valid ways to do that - but they weren't what my lecturer was referring too - they were discussed in the lecture itself. ). The way my lecturer asked this question was - " come up with ways to make several public keys related to a single private key - nevermind how vulnerable your solution is ".
@pg1989 has not only the right answer but the only answer if you assume the goal is to have multiple RSA encryption exponents that correspond to the same decryption exponent.

\begin{align}
e_1 \cdot d &amp; \equiv 1 \pmod{\phi(N)}\\
e_2 \cdot d &amp; \equiv 1 \pmod{\phi(N)}\\
(e_1 - e_2) \cdot d &amp; \equiv 0 \pmod{\phi(N)}
\end{align}

But since $d$ is relatively prime to $\phi(N)$, by Euclid's lemma $\phi(N)$ must divide $e_1 - e_2$ :

$$e_1 \equiv e_2 \pmod{\phi(N)}$$

In other words, any two public RSA exponents that share a private exponent must differ by some multiple of $\phi(N)$.

I highly suspect this is what your professor was thinking. Although, as @Thomas points out if the multiple is small and someone gets hold of both public keys, they can factor $N$. Now, maybe if you use a big multiple...

[Update]

As @fgrieu notes in a comment, I have conflated $\phi(N) = (p-1)(q-1)$ with $\operatorname{LCM}(p-1, q-1)$. Replace the former with the latter in the above derivation for the right answer.