Is it safer to encrypt twice with RSA? I wonder if it's safer to encrypt a plain text with RSA twice than it is to encrypt it just once. It should make a big difference if you assume that the two private keys are different, and that the only way used to crack it is brute force. I submitted these theories to my teacher, but he claims that a double encryption doesn't make it any safer. I didn't follow his arguments entirely, so I decided to ask here.

So, if I encrypt a message with one key once, and the encrypt the resulting cipher text once more with a different key, does this make the encryption simpler?

EDIT:
My teacher said that "it doesn't get safer with a double encryption, at least not if n is the same and e is different.". This is the part I don't follow, since you'd still need both p and q to derive the two different private keys that this would produce. I have made a few calculations and I do not quite understand. Why would the particular statement my teacher sent me mean that it doesn't get safer?
Well, think about it this way. If breaking one encryption with brute force will take longer than the lifetime of the universe, are you any safer with an encryption scheme that will take twice the lifetime of the universe? No. The first encryption cannot be broken. Adding a second encryption just adds computation overhead with no real benefit.

Think about it this way, if it is estimated to take 500 years for a prisoner to chew through the bars on his prison cell to escape, is the public any safer if we add a second set of bars so that it will take 1000 years to chew through the two sets before the prisoner can escape? Not really.

UPDATE
Given the update in the question, I thought I'd update.

So, you fix an $n$ and choose $e_1$ and $e_2$ as public exponents and compute $d_1$ and $d_2$ as the private exponents.

To encrypt, you are proposing $(m^{e_1})^{e_2}\bmod{n}$ and wondering why this is not stronger than just $m^{e_1}\bmod{n}$ in a brute-force attack[*].

So, you haven't given detail as to what the "brute-force" attack is, so let's look at two options.


Factoring $n$. If I factor $n$ using a brute-force attack, I then use the factorization to compute $d_1$ and $d_2$. Computing both $d_1$ and $d_2$ is not much more than just computing $d_1$ since you broke the factorization.
Instead of factoring $n$, what if you try to brute force $d_1$ and $d_2$. Recall that $d_i$ is chosen such that $e_i d_i\equiv 1\bmod{\varphi(n)}$. Furthermore, $(m^{e_1})^{e_2}=m^{e_1e_2}$. Raise that to $d_1d_2$ and you get $m$ back. Therefore, you really need to bruteforce $d_1d_2$ instead of $d_1$ and then $d_2$ (or vice-versa). If you assume each of the $d$s are $l$ bits, brute forcing $d_1$ then $d_2$ would be like brute forcing $l^2$ bits. Brute forcing $d_1d_2$ on the other hand is $2l$ bits. One could argue that this is harder, but asymptotically it isn't.
Brute force only $d_1$ then factor. It turns out if you know $d_1$ you can easily factor $n$ then use the factorization to compute $d_2$. (This comes from @CodesInChaos comment).


Any other brute force options you had in mind?

[*] My description of double encrypted RSA here is assuming textbook RSA. For padded RSA (which is what you find in the real world), points 1 and 3 are still valid, 2 however is not.