When you know $e,d,N$, you can calculate $ed-1$, which is a multiple of $\Phi(n)$. I guess that's what you meant by

$\Phi(n)$ will have multiple values.

The sentence itself is wrong, though. As a function it does not have "multiple values" for a fixed $n$. You know a multiple of the value.

There are various algorithms to do this:

• A probabilistic algorithm was given in the original RSA paper A method for obtaining digital signatures and public-key cryptosystems by Rivest, Shamir and Adleman, 1978
• A deterministic algorithm was given in Computing the RSA Secret Key is Deterministic Polynomial Time Equivalent to Factoring by May, 2004.
• This answer on SO references a different paper called Twenty Years of Attacks on the RSA Cryptosystem by Boneh, 1999.

This looks like a homework question, so I won't give an explicit algorithm.

When you know $e,d,N$, you can calculate $ed-1$, which is a multiple of $\Phi(n)$. I guess that's what you meant by

$\Phi(n)$ will have multiple values.

The sentence itself is wrong, though. As a function it does not have "multiple values" for a fixed $n$. You know a multiple of the value.

There are various algorithms to do this:

• A probabilistic algorithm was given in the original RSA paper A method for obtaining digital signatures and public-key cryptosystems by Rivest, Shamir and Adleman, 1978
• A deterministic algorithm was given in Computing the RSA Secret Key is Deterministic Polynomial Time Equivalent to Factoring by May, 2004.
• This answer on SO references a different paper called Twenty Years of Attacks on the RSA Cryptosystem by Boneh, 1999.

This looks like a homework question, so I won't give an explicit algorithm.