I'm currently studying cryptography and have the following question.
Suppose we have the following relation in an RSA system $ed\equiv1\pmod{\Phi(pq)}$ and $ek\equiv1\pmod{\lambda(pq)}$ Show that both $d$ and $k$ can be used to decrypt.
How can I show this? I know that $\lambda | \Phi$ and $d\equiv k\pmod\lambda$ but why does this help us decrypt using either $d$ or $k$.
On
If $k$ is such that $ek=1 \pmod{\lambda(n)}$ (for some number $k \in \{1,\dots,\lambda(n)-1\}$), then $k$ works as an inverse for encryption exponent $e$ in RSA, because we know that if $x$ is some "message", relatively prime to $n$, by definition of $\lambda$ (!) we know that $x^{\lambda(n)}=1 \pmod{n}$ and then writing $ek-1 = N\cdot\lambda(n)$ for some $N \in \mathbb{Z}$, we know that $$(x^e)^k = x^{ek} = x^{1+N\lambda(n)} = x^1 \cdot (x^{\lambda(n)})^N = x \pmod{n}$$
Now if $ed=1 \pmod{\phi(n)}$ for some $d \in \{1, \ldots, \phi(n)-1\}$ we know that $\phi(n) | (ed-1)$ and as $\lambda(n)|\phi(n)$, $\lambda(n)|(ed-1)$ as well, and so $d$ (or $d \pmod{\lambda}$ if $d > \lambda(n)$) will work by the above.
It's slightly more efficient to work modulo $\lambda(n)$ as it is smaller than $\phi(n)$ in the RSA situation (typically, the way the parameters are chosen, we have $\lambda(n)=\frac{\phi(n)}{2}$, as $\gcd(p-1,q-1)=2$ in a typical setup.) Most standards even proscribe working modulo $\lambda(n)$.
If any $k$ such that $ek\equiv1\pmod{\lambda(pq)}$ can be used, then any $d$ such that $ed\equiv 1\pmod{\varphi(pq)}$ can, because $\lambda(pq)\mid\varphi(pq)$ as you noted, and then $ed\equiv 1\pmod{\lambda(pq)}$.
To prove that $k$ can be used means to prove that $x^{ek}\equiv x\pmod{pq}$ for all $x$; in fact, more generally, for any squarefree $n$, and any $m>0$ such that $m\equiv 1\pmod{\lambda(n)}$, we have $x^m\equiv x\pmod{n}$ for all $x$ (indeed, $n$ is a product of distinct primes, and for any such prime $p$ we have $x^m\equiv x\pmod{p}$, because $\color{gray}{p-1={}}$$\lambda(p)\mid\lambda(n)$${}\mid(m-1)$ — so the claim follows by CRT).