Prove that $n$ is prime iff $\sigma(n)+\phi(n)= nd(n)$. I am sorry if this post is a duplicate, but I could not seem to find a proof following this method in my searches and, while I think my reasoning is sound, I’m not sure if my inductive proof is rigorously complete. 
In the following $\sigma(n)$, $\phi(n)$, and $d(n)$ are, respectively, the sum of the divisors of $n$, the Euler totient function, and the  total number of divisors of $n$. I have broken the proof into several Lemmas:

Lemma 1: If $n$ is a prime number, then $$\sigma(n)+\phi(n)=nd(n).$$

Proof:
This follows directly from definitions, i.e.:
$$\sigma(n)+\phi(n)=(1+n)+(n-1)=2n=n d(n).$$$\square$

Lemma 2: If $n=p^a$, where $p$ is any prime and $a$ is an integer greater than $1$, then $$\sigma(n)+\phi(n)\lt n d(n).$$

Proof:
Assume, for a contradiction, that $$\sigma(n) +\phi(n)\geq nd(n),$$ this is equivalent to $$\left(1+p+\cdots +p^a\right) +\left(p^a-p^{a-1}\right)\geq p^a(a+1)\\\implies 1+p+\cdots +p^{a-2}\geq (a-1) p^a,$$ which, since there are $a-1$ terms in the sum on the LHS of the last inequality, each less than $p^a$, is the contradiction we seek.
$\square$

Lemma 3: If $$n={p_1}^{a_1}{p_2}^{a_2}\cdots {p_k}^{a_k},$$ where the $p_i $ are all distinct primes and $a_i \geq1$, then $$\sigma(n)+\phi(n)\leq nd(n),$$ with equality if and only if $k=1$, $a_1=1$.

We prove this by induction. The case $k=1$ follows from Lemmas 1 and 2. For the inductive hypothesis, assume that, for any collection of $k$ distinct primes, $k\in\Bbb{N}$, it is always true that, if $$n={p_1}^{a_1}{p_2}^{a_2}\cdots {p_k}^{a_k},$$ then $$\sigma(n)+\phi(n)\leq nd (n).$$
Now, let $$m={p_{k+1}}^{a_{k+1}},$$ such that $gcd(m,n)=1,$ then
$$\sigma(mn)+\phi(mn)=\sigma(m)\sigma(n)+\phi(m)\phi(n)\\\leq\sigma(m)\big[nd(n)-\phi(n)\big]+\phi(m)\big[nd(n)-\sigma(n)\big]\\=nd (n)\big[\sigma(m)+\phi(m)\big]-\big[\sigma(m)\phi(n)+\phi(m)\sigma(n)\big]\\\lt n d(n)\big[\sigma(m)+\phi(m)\big]\\\leq mn\big[ d(mn)\big],$$ where the first and last inequalities follow from Lemmas 1 and 2.
$\square$
 A: As you mentioned in a comment on my (now deleted) answer, you haven't established the strict inequality that Lemma 3 requires. The base case is fine, as Lemma 2 shows a strict inequality for non-prime numbers covered by the $k = 1$ case. But, you haven't showed a strict inequality for $k > 1$.
You could cover most cases, by simply insisting that $a_1 \ge a_i$ for all $i = 1 \ldots, k$, which can be achieved without loss of generality by rearranging the primes. Then, if $a_i > 1$, the strict inequality from the base case would carry over inductively. But, this still leaves you with square-free numbers of the form $p_1 p_2 \ldots p_k$.
What you need to do is add in another lemma, showing that $\sigma(n) + \phi(n) < nd(n)$ for $n = p_1 p_2$, where $p_1 \neq p_2$ are primes. It's not hard to show:
$$\sigma(n) + \phi(n) = (1 + p_1)(1 + p_2) + (p_1 - 1)(p_2 - 1) = 2p_1p_2 + 2 < 4p_1 p_2.$$
Now, in terms of constructing your induction argument, I'd treat the $k = 1$ case as totally separate. Then, prove the base case $k = 2$, but keep the inequality strict. The two above arguments show that the inequality is strict for $n = p_1^{a_1}p_2^{a_2}$ regardless of whether $n$ is square-free or not. Then, your induction step will work as is, just with strict inequalities, instead of non-strict inequalities.
