What are the minimum constraints on RSA parameters and why?
One does not need ​ e &lt; n .

One does not need $\;\; e\hspace{-0.03 in}\cdot \hspace{-0.03 in}d \: \equiv \: 1 \;\; \pmod{\phi(n)} \;\;\;\;\;$;

one just needs that congruence modulo λ(n).

(Being coprime with $\phi$(n) is equivalent to being coprime with λ(n).)

Neither the plaintext nor the ciphertext need to be coprime with n.

gcd(m,n) must be coprime with n/(gcd(m,n)).



That last condition will be automatic when n is square-free, since it's equivalent to

"If m≠0 then for every prime factor p of n, p's multiplicity as a

factor of m is either 0 or equal to p's multiplicity as a factor n.".

When the quote holds, for each maximal-prime-power factor q of n,

m's remainder mod q will be either 0 or a coprime to q.

When the former case applies, it will continue applying to all positive powers of m.

When the latter case applies:

By Fermat's Little Theorem, $\;\;\; m^{\hspace{.02 in}\phi(q)} \equiv 1 \: \pmod{q} \:\:\:\:$.

For all non-negative integers k,

$m^{(k\cdot \phi(q))\hspace{.02 in}+1} \: \equiv \: m^{\hspace{.02 in}k\cdot \phi(q)} \cdot m^1 \: = \: m^{\hspace{.02 in}\phi(q) \cdot k} \cdot m \: =$

$\left(\hspace{-0.03 in}m^{\hspace{.02 in}\phi(q)}\hspace{-0.06 in}\right)^k \cdot m \: \equiv \: 1^k \cdot m \: = \: 1\cdot m \: = \: m \;\;\; \pmod{q} \;\;\;\;\;\;\;\;$.

λ(n) is a positive integers multiple of $\phi$(q), so for all

non-negative integers $k$, $\;\;\; m^{(k\cdot \lambda(n))\hspace{.02 in}+1} \equiv m \: \pmod{q} \:\:\:\:$.

Thus, when the quote holds, one will have $\;\;\; m^{(k\cdot \lambda(n))\hspace{.02 in}+1} \equiv m \: \pmod{q}$

for each maximal-prime-power factor q of n and each non-negative integer k.

Therefore, by the Chinese Remainder Theorem, when the quote holds one

will have $\;\;\; m^{(k\cdot \lambda(n))\hspace{.02 in}+1} \equiv m \: \pmod{n} \;\;\;$ for all non-negative integers k.