RSA decryption needs Euler, and Euler needs a coprime message

Let $p=5$ and $q=13$, and set $n=pq=65$. Let $e=5$. 1. Compute $\varphi(n)$ from the product formula, and prove that $\gcd(e,\varphi(n))=1$. Find the unique $d$ with $1\le d<\varphi(n)$ and \[ ed\equiv 1\pmod{\varphi(n)}. \] Do not invert $e$ modulo $n$. 2. Prove Euler's theorem: if $\gcd(a,n)=1$, then $a^{\varphi(n)}\equiv 1\pmod n$. Your proof should use that multiplication by $a$ permutes $(\mathbb Z/n\mathbb Z)^\times$. 3. Let $m$ be an integer with $\gcd(m,n)=1$, and set $c\equiv m^e\pmod n$. Using only Euler's theorem and the congruence $ed\equiv 1\pmod{\varphi(n)}$, prove \[ c^d\equiv m\pmod n. \] Then, for the plaintext $m=11$, compute the ciphertext $c\equiv 11^e\pmod n$ and verify $c^d\equiv 11\pmod n$. Record both residues in $\{0,1,\ldots,n-1\}$. 4. Let $m'=10$. Compute $\gcd(m',n)$ and prove that Euler's theorem does not apply to $m'$ modulo $n$, by showing $ (m')^{\varphi(n)}\not\equiv 1\pmod n$. Explain why quoting Euler on $m'$ cannot justify RSA decryption of $m'$, even if some other argument (not requested here) might still recover $m'$. Also explain why the private exponent must be an inverse of $e$ modulo $\varphi(n)$ rather than modulo $n$. Do not treat $n$ as prime, do not replace $\varphi(n)$ by $n-1$, and do not skip the coprime-message hypothesis.

$\varphi(65)=4\cdot 12=48$. The inverse of $5$ modulo $48$ satisfies $5d=48k+1$; try $k=3$.