# Does the following equation have any solutions in $\mathbb{N}$?

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 $$x \in \mathbb{N}$$.

• Maybe you can utilize a formula form the following link:math.stackexchange.com/questions/22721/… – NoChance Oct 13 '15 at 8:17
• Thank you very much for pointing me to that MSE link, @NoChance! – Arnie Bebita-Dris Oct 13 '15 at 8:23
• 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. – mathlove Nov 6 '18 at 11:07
• I would be more than delighted to hear about your results, @mathlove. =) – Arnie Bebita-Dris Nov 6 '18 at 11:31
• 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. – Ross Millikan Jan 4 at 21:36

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 $$i$$.$$\quad\square$$
• Just a minor comment: Surely $3 \not\mid 3x+1$ as $3x+1 \equiv 1 \pmod 3$? =) – Arnie Bebita-Dris Nov 6 '18 at 11:42