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Let $\mathbb{N}$ be the set of positive integers.

The function $\sigma(N)$ gives the sum of the divisors of $N$.

My question is:

Does the following equation have any solutions for $x \in \mathbb{N}$? $$\sigma(3x + 1) = 4x + 1$$

Notice that $3x + 1$ must be deficient, and that if we allow $x = 0$, then it is a (trivial) solution.

Furthermore, suppose that $3x + 1 = p^{\alpha}$, where $p$ is prime. Then we have $$\sigma(3x + 1) = \frac{p^{\alpha+1} - 1}{p - 1} = \frac{4(p^{\alpha} - 1)}{3} + 1 = 4x + 1,$$ from which we obtain $$3(p^{\alpha+1} - 1) = 4(p - 1)(p^{\alpha} - 1) + 3(p - 1).$$ Simplifying and collecting like terms, we get $$0 = p^{\alpha+1} - 4p^{\alpha} + p + 4.$$ Rewriting the last equation, we have $$(p^{\alpha} + 1)(4 - p) = 8.$$ As $p^{\alpha} + 1 \geq 3$, we have the possibilities $$(p^{\alpha} + 1, 4 - p) \in \{(4,2),(8,1)\}.$$ Both of them are impossible under the given constraints.

Consequently, $\omega(3x + 1) \geq 2$, where $\omega(y)$ is the number of distinct prime factors of $y$. (That is, $3x + 1$ must be composite.)

Lastly, I tried checking for equality (in the range $0 \leq x \leq 100$) using a spreadsheet, and found only the solution for $x=0$.

I am therefore compelled to predict that:

The equation $\sigma(3x + 1) = 4x + 1$ does not have any solutions for $x \in \mathbb{N}$.

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    $\begingroup$ Maybe you can utilize a formula form the following link:math.stackexchange.com/questions/22721/… $\endgroup$ – NoChance Oct 13 '15 at 8:17
  • $\begingroup$ Thank you very much for pointing me to that MSE link, @NoChance! $\endgroup$ – Arnie Bebita-Dris Oct 13 '15 at 8:23
  • $\begingroup$ Some small results I've got are (1) $2\not\mid 3x+1$ (2) $3\not\mid 3x+1$ (3) $3x+1$ has to be a perfect square. $\endgroup$ – mathlove Nov 6 '18 at 11:07
  • $\begingroup$ I would be more than delighted to hear about your results, @mathlove. =) $\endgroup$ – Arnie Bebita-Dris Nov 6 '18 at 11:31
  • $\begingroup$ It would be better just to say you have checked over a range instead of listing the whole spreadsheet. It takes up a lot of space needlessly. $\endgroup$ – Ross Millikan Jan 4 at 21:36
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Claim 1 : $2\not\mid 3x+1$.

Proof : $x=1$ is not a solution.

For $x\ge 2$, suppose that $2$ is a divisor of $3x+1$. Then, we have $$4x+1=\sigma(3x+1)\ge 1+(3x+1)+2+\frac{3x+1}{2}$$ There are no $x\ge 2$ satisfying the inequality.$\quad\square$

Claim 2 : $3x+1$ is a perfect square.

Proof : Let $3x+1={p_1}^{a_1}{p_2}^{a_2}\cdots {p_k}^{a_k}$ where $p_1,p_2,\cdots, p_k$ are distinct primes larger than $3$ and $a_i\ge 1\in\mathbb Z$. Then, the equation becomes $$(1+\cdots +{p_1}^{a_1})(1+\cdots +{p_2}^{a_2})\cdots (1+\cdots +{p_k}^{a_k})=4x+1$$The RHS is odd since $x$ is even. So, $a_i$ has to be even for every $i$.$\quad\square$

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  • $\begingroup$ Just a minor comment: Surely $3 \not\mid 3x+1$ as $3x+1 \equiv 1 \pmod 3$? =) $\endgroup$ – Arnie Bebita-Dris Nov 6 '18 at 11:42
  • $\begingroup$ @Jose Arnaldo Bebita Dris : You are right:) Deleting it. $\endgroup$ – mathlove Nov 6 '18 at 11:43

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