# Deductions about the prime factorization of odd integers $m>1$ satisfying $\frac{-3+\sqrt{1+8m}}{2}=\prod_{\substack{p\mid m\\p\text{ prime}}}(p-1)$

For each integer $m>1$ we denote the Euler's totient function as $\varphi(m)$ and the product of distinct primes dividing it, that is its square-free kernel, as $\operatorname{rad}(m)$ (that is the arithmetic function explained in this Wikipedia). Thus for our integer $m>1$ one has $$\varphi(\operatorname{rad}(m))=\prod_{\substack{p\mid n\\p\text{ prime}}}(p-1).$$

While I was calculating solutions of some equations that involves* the arithmetic function $\varphi(\operatorname{rad}(n))$ and the sum of divisors and totatives for even perfect numbers, I got also odd solutions of next equation $$\frac{-3+\sqrt{1+8n}}{2}=\varphi(\operatorname{rad}(n)).\tag{1}$$

The odd solutions of $(1)$ that I can to calculate are $n=45$ and $n=3321$. These solutions have the form $q\cdot t^2$, being $q$ an odd prime $\equiv 1\text{ mod }4$ and $\gcd(q,t)=1$, in fact in previous solutions $t$ is a power of $3$. I would like to know what work can be done about the characterization of such odd solutions, if there are more odd integers being solutions.

Question. Imagine that we need to study the odd integers $m>1$ satisfying $(1)$. What can we deduce about such odd solutions $m$? I am asking about their prime factorization. Many thanks.

One has thus that we need to study the prime factorization of odd integers $m$ that satisfy $$\frac{-3+\sqrt{1+8m}}{2}=\prod_{\substack{p\mid m\\p\text{ prime}}}(p-1).\tag{2}$$

• *The equations that I evoke (thus comment this isn't directly related to previous $(1)$ and the Question), and that satisfy each even perfect numbers $e$ are this $\frac{-3+\sqrt{1+4\sigma(e)}}{2}=\varphi(\operatorname{rad}(e))$ being $\sigma(n)$ the sum of divisors function, and thisone $2(e-1)-\varphi(e)=\left(\frac{11+3\sqrt{1+8e}}{8}\right)\varphi(\operatorname{rad}(e))$. I add this comment as context, and as a comparison of previous equations, $(1)$ is the simplest equation, but there exist (odd) integers satisfying it that aren't even perfect numbers. – user243301 May 10 '18 at 14:08

Calculation shows that$$\frac{-3+\sqrt{1+8m}}{2}=\phi(rad(m))$$for odd $m= $$45$$$$3321$$$$926510115949281$$$$1716841910146257168839018970561$$These numbers have the form $$\frac{3^n(3^n+1)}{2}$$for even$n=2, 4, 16, 32$, respectively. Note that$m$is thus the$3^nth$triangle number. Indeed,$\sqrt{1+8m}$is an integer only if odd$m$is triangular, since for integer$k$$\sqrt{1+\frac{8k(k+1)}{2}}=\sqrt{4k^2+4k+1}=\sqrt{(2k+1)^2}=2k+1$$

Factoring the four $m$:$$45=3^2(3^2+1)/2=3^2\cdot 5$$$$3321=3^4(3^4+1)/2=3^4\cdot 41$$$$926510115949281=3^{16}(3^{16}+1)/2=3^{16}\cdot 21523361$$$$1716841910146257168839018970561=3^{32}(3^{32}+1)/2=3^{32}\cdot 926510094425921$$For even $n$ from $2$ through $32$, only these four $m$ have the form $3^np$, i.e are composed of a power of $3$ and a single odd prime $p$.

This condition appears to be sufficient for$$\frac{-3+\sqrt{1+8m}}{2}=\phi(rad(m))$$and can be proven as follows:

If $$m=3^np$$ then $$rad(m)=3p$$and$$\phi(rad(m))=(3-1)(p-1)=2p-2$$

And again, if$$\sqrt{1+8m}=\sqrt{1+8\cdot \frac{3^n(3^n+1)}{2}}=\sqrt{1+8\cdot \frac{3^{2n}+3^n}{2}}=\sqrt{4\cdot 3^{2n}+4\cdot 3^n+1}=2\cdot 3^n+1$$then$$\frac{-3+\sqrt{1+8m}}{2}=\frac{-3+2\cdot 3^n+1}{2}=3^n-1$$And since$$p=\frac{3^n+1}{2}$$then$$3^n-1=2p-2$$as above, and$$\frac{-3+\sqrt{1+8m}}{2}=\phi(rad(m))$$for all odd $m=3^np$.

Note that $n$ must be even, since if $n$ is odd, then $3^n+1=0(mod4)$, making $\frac{3^n+1}{2}$ even and hence not an odd prime.

At first, I expected the equation might hold for all $n$ equal to a power of $2$, but $n=8$ does not work. Primes are infinitely many, but if we cannot know the law of even $n$ for which $(3^n+1)/2$ is prime, then even if $$\frac{-3+\sqrt{1+8m}}{2}=\phi(rad(m))$$is true for infinitely many odd $m$, it seems that, like the primes themselves, we will have to find them one at a time.

Further calculation reveals that $\frac{3^n+1}{2}$= $$1716841910146256242328924544641$$is prime for $n=64$, when $m$=$$5895092288869291585760436430707976174749252052380249479093121$$But $\frac{3^n+1}{2}$ is not prime for $n=128, 256, 512$ or $1024$.

More importantly, it appears--and I suppose it can be proven--that $5$ is a divisor of $\frac{3^n+1}{2}$ not only for $n=2$ but for every $n$ which is an odd multiple of $2$. And since all these increasingly large $\frac{3^n+1}{2}$ must contain other prime factors besides $5$, they cannot be prime. This rules out $$n=6, 10, 14, 18, 22...$$or one half of all even $n$, less one.

Similarly, $41$ is a divisor of $\frac{3^n+1}{2}$ for $n=4$ and all odd multiples of $4$, so that $\frac{3^n+1}{2}$ cannot be prime when $n$ is an odd multiple of $4$. This rules out half of the remaining even $n$, less one$$12, 20, 28, 36, 44...$$

Again, since $17$ and $193$ divide $\frac{3^n+1}{2}$ for $n=8$, as well as for all odd multiples of $8$, then$$8, 24, 40, 56, 72...$$or exactly half of the remaining $n$ are ruled out.

Accordingly, since $n$ is even, and an even number which is not an odd multiple of $2, 4, 8, 16, 32 ...$ must be a power of $2$, it seems a necessary but insufficient condition of the primality of $\frac{3^n+1}{2}$ that $n$ be a power of $2$.

Hence, like the Fermat primes, the odd $m$ for which$$\frac{-3+\sqrt{1+8m}}{2}=\phi(rad(m))$$are not only rare, but perhaps finite in number.

• Many thanks for your attention, calculations and reasonings. – user243301 May 15 '18 at 20:47
• @user243301 If you want, you can $\color{green}{\text{accept}}$ this answer :) – Mr Pie May 16 '18 at 1:12