On the equation $\varphi(n)=\left(\frac{1+\sqrt{1+8n}}{8}\right)\cdot\left(\operatorname{rad}(n)-\frac{1+\sqrt{1+8n}}{2}\right)$

Question

An integer is said to be an even perfect number if satisifies $\sigma(n)=2n$, where $\sigma(n)$ is the sum of the positive divisors of $n$. The first few even perfect numbers are $6,28,496$ and $8128$.

Question. I wondered if we can prove or refute the following statement:

Let $\varphi(n)$ the Euler's totient function and we denote the product of the distinct primes dividing a natural $n>1$ as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p$$ (it is this arithmetic function from the Wikipedia Radical of an integer). If $n$ satisfies the equation $$\varphi(n)=\left(\frac{1+\sqrt{1+8n}}{8}\right)\cdot\left(\operatorname{rad}(n)-\frac{1+\sqrt{1+8n}}{2}\right)$$ then $n$ is an even perfect number.

Many thanks.

Answer 1

The converse seems to hold: If $n$ is an even perfect number, then the equation is true.

At the end of Elements IX, Euclid shows that if $p$ and $2^p-1$ are prime, then$$n=\frac{(2^p-1)2^p}{2}$$is a perfect number.

Granting that consequently$$\phi(n)=\frac{(2^{p-1}-1)2^p}{2}=(2^{p-1}-1)2^{p-1}$$and that$$rad(n)=(2^p-1)2$$then by substitution in the given equation$$(2^{p-1}-1)2^{p-1}=\frac{1+\sqrt{1+8\frac{(2^p-1)2^p}{2}}}{8}\cdot [(2^p-1)2-\frac{1+\sqrt{1+8\frac{(2^p-1)2^p}{2}}}{2}]$$And since it can be shown that$$\frac{1+\sqrt{1+8\frac{(2^p-1)2^p}{2}}}{8}=2^{p-2}$$and$$\frac{1+\sqrt{1+8\frac{(2^p-1)2^p}{2}}}{2}=2^p$$then again by substitution$$(2^{p-1}-1)2^{p-1}=2^{p-2}[(2^p-1)2-2^p]$$Hence by distribution$$2^{2p-2}-2^{p-1}=2^{2p-1}-2^{p-1}-2^{2p-2}$$and eliminating and transposing$$2^{2p-2}+2^{2p-2}=2^{2p-1}$$that is$$2(2^{2p-2})=2^{2p-1}$$or$$2^{2p-1}=2^{2p-1}$$

Answer 2

Computational results:

Your equation requires $\sqrt{1+8n} \equiv 3 \mod 4$. This holds when $n = 2m^2 + 3m + 1, \ m \in \mathbb Z$.

I have used Mathematica to test $m \le 10^6$, or $n \le 2 \times 10^{12}$. The only solutions are the first 7 perfect numbers.

rad[n_] := Times @@ First /@ FactorInteger[n];
f[n_] := ((1 + Sqrt[1 + 8 n])/8) (rad[n] - (1 + Sqrt[1 + 8 n])/2);
Select[Table[2 m^2 + 3 m + 1, {m, 1, 10^6}], EulerPhi[#] == f[#] &]

Answer 3

Using $n = m(2m-1)$ and solving for $\varphi(n)$: $$\varphi(n) = \frac m 2 \operatorname{rad}(n) - m^2$$

Since $2m-1$ and $m$ are coprime, and $\operatorname{rad}$ and $\varphi$ are multiplicative, we have $$\varphi(2m-1)\varphi(m) = \frac m 2 \operatorname{rad}(2m-1) \operatorname{rad}(m) - m^2$$

Now we write $m = 2^k d$ for some odd $d$. $$\varphi(2m-1) \varphi(2^k d) = \frac m 2 \operatorname{rad}(2m-1) \operatorname{rad}(2^k d) - m^2$$

$$\varphi(2m-1) \varphi(d)2^{k-1}=m \operatorname{rad}(2m-1) \operatorname{rad}(d) - m^2$$

All that is left is to prove or disprove $d=1$ and $2m-1$ is prime to conclude $n$ is a perfect number.

Observations:

The RHS has a factor of $m$, and so has a factor of $d$, and since $\varphi(d) < d$ then we must have $d \mid \varphi(2m-1)$.

If $d=1$ then $\varphi(2m-1) = 2 \operatorname{rad}(2m-1) - 2m$. As far as I can tell this means $2m-1$ is prime.

If $d$ is prime then we have $\varphi(2m-1) (d-1) 2^{k-1}= md \operatorname{rad}(2m-1) - m^2$. RHS has a factor of $d^2$ so we must have $d^2 \mid \varphi(2m-1)$.

If $2m-1$ is prime then $(2m-2) \varphi(d)2^{k-1} = m (2m-1) \operatorname{rad}(d) - m^2$. Using $2^{k-1} = \frac m {2d} $ this is $(m-1)\varphi(d) = d((2m-1)\operatorname{rad}(d) - m)$. Not sure if this helps