This problem has a one liner solution. But there is another beautiful solution too, though that may be frustratingly a large one. I hope that you find both. Specially because, the latter solution can be used as a tactic in many problems.
Prove that, for a positive integer $k$, the following equation has only finitely many solutions:
\[\sigma(n)=n+k\]
Here $\sigma(n)$ is the sum of divisors of $n$.
Equation On Sum Of Divisor
One one thing is neutral in the universe, that is $0$.
Re: Equation On Sum Of Divisor
Hint/ideas for the second solution? Because after finding the one liner, there are too many ways to take on this.
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Re: Equation On Sum Of Divisor
Let $S=\{n\in\mathbb N:k=\sigma(n)-n\}$ and $T=\{p\in \mathbb P:\exists n\in S:p|n, n\in S\}$ and $E=\{e\in\mathbb N:\exists p\in T:p^e|||n\in S\}$. Now, show that $E$ and $T$ are finite.
One one thing is neutral in the universe, that is $0$.