RSA KeyPair generation with Bouncy Castle provider: $e*d(φ(n))$ not equal to one
As CodesInChaos notes in the comments, the necessary and sufficient condition for $m^{ed} \equiv m \pmod{n}$ to hold for all messages $m$ is that $ed \equiv 1 \pmod{\lambda(n)}$, where $\lambda(n) = \operatorname{lcm}(p-1, q-1)$ is the Carmichael reduced totient function of the modulus $n = pq$.

As it happens, the Euler totient $\varphi(n) = (p-1)(q-1)$ is an integer multiple of $\lambda(n)$, so if $ed \equiv 1$ modulo $\varphi(n)$, then this also holds modulo $\lambda(n)$.  Thus, when generating an RSA key, it's OK to compute the private exponent $d \equiv e^{-1}$ modulo $\varphi(n)$ rather than modulo $\lambda(n)$, and indeed quite a few introductory texts on RSA (including, apparently, Wikipedia) do that for some reason, perhaps because it's arguably a bit easier to explain the math that way, or perhaps simply because the authors also mistakenly think that the necessary condition for key validity is $ed \equiv 1 \pmod{\varphi(n)}$.

However, this is not actually the case, and it's indeed quite possible for $(n,e,d)$ to be a valid RSA key even if $ed \not\equiv 1 \pmod{\varphi(n)}$, as long as $ed \equiv 1 \pmod{\lambda(n)}$.  Since using $\lambda(n)$ also yields the smallest valid (positive) private exponent $d$ for any given $e$ and $n$, and since computing $\lambda(n)$ is not significantly more difficult than computing $\varphi(n)$, most practical RSA implementations indeed do just that.