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If $$n=p_1^{a_1}\cdots p_k^{a_k},$$ then define

$$f(n):=p_1^2+\cdots+p_k^2$$

So, $f(n)$ is the sum of the squares of the prime divisors of $n$.

For which natural numbers $n\ge 2$ do we have $f(n)=n$ ?

It is clear that $f(n)=n$ is true for the square of any prime, but false for the other prime powers.

If $p$ and $q$ are the only prime divisors of $n$, we would get $p^2+q^2\equiv 0\pmod p$, which implies $p=q$, so for numbers with exact two prime divisors, $f(n)=n$ cannot hold.

If $p,q,r$ are primes with $p<q<r$, then we have two possibilities.

If $p,q,r\ne 3$, we have $p^2+q^2+r^2\equiv 0\pmod3$, so $f(n)=n$ cannot hold. If $p=3$ or $q=3$, then $p^2+q^2+r^2 \equiv 2\pmod3$, so $p^2+q^2+r^2$ is not divisible by $3$, so $f(n)=n$ cannot hold.

Finally, if $p<q<r<s$, then if $p>2$, then $p^2+q^2+r^2+s^2\equiv 0\pmod4$, so $f(n)=n$ cannot hold. And if $p=2$, then $p^2+q^2+r^2+s^2\equiv 3\pmod4$, so $p^2+q^2+r^2+s^2$ is odd and $f(n)=n$ again cannot hold.

So, apart from the squares of the primes, the number must have at least $5$ prime factors. I searched to about $6\times 10^7$ and did not find a "non-trivial" example.

  • Is there a number $n$ with at least two prime factors and $f(n)=n$ ?
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  • 2
    $\begingroup$ Similar question (but different of course). $\endgroup$ – user26486 Jun 2 '15 at 17:33
  • $\begingroup$ I found no examples for $n\le 10 ^9$. $\endgroup$ – Peter Jun 2 '15 at 18:17
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    $\begingroup$ A small remark (again): suppose $p_1\lt\cdots\lt p_m.$ Then $$mp_m^2\ge p_1^2+\cdots+p_m^2=p_1^{a_1}\cdots p_m^{a_m}\ge m!p_m^{a_m}.$$ Hence $a_m\le2.$ If $a_m=2,$ then $m\ge m!$ implies that $m=1,2,$ already discussed by you. So $a_m=1.$ But I failed to extend this argument to obtain satisfying results. $\endgroup$ – awllower Jun 3 '15 at 0:27
  • $\begingroup$ You should have mentioned a proof of $p_1^{a_1}...p_{m-1}^{a_{m-1}}\ge m!$,but in fact, even the product of the first $m-1$ primes is at least $m!$ $\endgroup$ – Peter Jun 3 '15 at 10:01
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    $\begingroup$ Well, it seems you already have a proof. ;-) $\endgroup$ – awllower Jun 3 '15 at 23:32
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Sorry because this isn't a full answer, but I believe that contains sustancials statments that could provide an improve from someone.

Let $d|n$, thus $n=d\cdot n/d$, and by symmetry $\prod_{d|n}d=\prod_{d|n}n/d$, multiply $\prod_{d|n}$ in first identity states $$\left( \prod_{d|n}d\right)^{2}=n^{\sigma_{0}(n)}$$ (this is Exercise 10, page 47 from Apostol, Introduction to Analytic Number Theory), where $\sigma_{0}(n)$ is the number of divisors function. My attempt is extract arithmetic information from this and Euler-Fermat Theorem. The following cases are disjoint with fullness for a collection of primes.

Case 1. We assume without loss of generality that the first prime is $2$, we obtain ($n>1$) $$\left( p_{1}^{2}+p_{2}^{2}+\cdots +p_{\omega (n)}^{2}\right)^{\sigma_{0}(n)}=n^{\sigma_{0}(n)}=\prod_{d|n}d^{2},$$ thus $(0+\omega (n)-1)^{\sigma_{0}(n)}\equiv 0\mod 4$, where $\omega (n)$ is the number of distinct primes, since if $m$ is odd, $m^{2}\equiv 1\mod 4$. These computations removes the cases $\omega(n)$ equals $4\lambda$ or $4\lambda +2$, there are infinitely many subcases.

Case 2. All primes are odd, we obtain with same idea $\omega (n)^{\sigma_{0}(n)}\equiv 1\mod 4$ and discard the same subcases (caution, removes subcases in this case).

Perhaps bounding or using another identities someone can to sweep more cases. I understand that this isn't the full answer, so I accept the response of community, yours or moderators.

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  • $\begingroup$ Thanks for vote up my partial answer. $\endgroup$ – user243301 Nov 8 '15 at 7:30
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It's not a full answer, just remarks

  • First

$a_k=1$ (as stated by Awllower)

$a_{k-1} = 1$

$k$ can't be even (obvious)

As stated by Peter $k$ can't be 3, and for the same reason $k$ can't be a multiple of 3, and for the same reason if $k=3x+2$ then 3 can't be a factor of $n$.

  • Then

set $x=k$

we have $x*p_k^2>n$

<=> $x*p_k>\prod_{i=1}^{k-1} p_i^{a_i}$

<=> $x*p_k/\prod_{i=1}^{k-2} p_i^{a_i}>p_{k-1}$

so $p_k^2+(k-1)*(x*p_k/\prod_{i=1}^{k-2}p_i^{a_i})^2>n$

<=> $p_k^2*(1+(k-1)*(x/\prod_{i=1}^{k-2}p_i^{a_i})^2)>n$

Now if we try to get the lowest value for $x$, the limite is :

$x=1+(k-1)*(x/\prod_{i=1}^{k-2} p_i^{a_i})^2$

<=> $x={\prod_{i=1}^{k-2}p_i^{a_i}*(\prod_{i=1}^{k-2}p_i^{a_i}-\sqrt{(\prod_{i=1}^{k-2}p_i^{a_i})^2-4(k-1)})\over2(k-1)}$

numerical evaluation of $x$ for $k=5$ and $\prod_{i=1}^{k-2}p_i^{a_i}=2*5*7$ gives $x<1,0008177$

Hope it can help.

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Not an answer, but another point of view

$\prod_{i=1}^{k}p_i^ai=\sum_{i=1}^kp_i^2=(\prod_{i=1}^{k}p_i^{ai-1})*(\prod_{i=1}^{k}p_i)$

So it's a Hurwitz equation (More complete but in French)

Lets call $A=\prod_{i=1}^{k}p_i^{ai-1}$

If there is a solution, for $k\gt2$ and $p_i\neq0$, we have $1 \leq A \leq k.$

We can put some restrictions on fundamentales solutions due to the fact that $p_1,p_2,...,p_k$ are distinct primes :

A fundamental solution can't contain coprimes.

A fundamental solution must contain all distinct primes of $A$.

There is no solutions if $k \equiv 0\bmod2$ or if $k \equiv 0\bmod3$.

The fundamental solutions for $k\leq10$ are (source post of the image) :

Hurwitz table

So for $k\leq10$ the only possible fundamental solutions are :

$k=5,A=4,(1,1,1,1,2)$

$k=7,A=3,(1,1,1,1,1,2,3)$

I don't have a software to explore these fondamental solutions, but they may lead to a solution to the main question.

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As awllower noted in comments, if $p_1<p_2<\ldots < p_k$, then $a_k=1$.

Then, based on this property, we rewrite numbers $n$ and $f(n)$ in the form: $$ n = q\cdot p_k, \quad \mbox{ where} \quad q = p_1^{a_1}\cdots p_{k-1}^{a_{k-1}};\tag{1} $$ $$ f(n) = p_k^2+f(q).\tag{2} $$ Then we focus on equation $$ p_k^2+f(q) = q\cdot p_k, $$ which is quadratic equation (for variable $p_k$): $$ p_k^2 - q\cdot p_k + f(q) = 0.\tag{3} $$ Then we require next conditions:
a) $q^2-4f(q) = D^2$, where $D \in\mathbb{N}$;
b) $\frac{q-D}{2}$ or $\frac{q+D}{2}$ is prime number.

This method allows to focus on smaller factors of a number $n$ and construct (?)(if possible) the largest appropriate prime factor.

But strange that I cannot find any such $q$ that at least condition a) holds: $$ q^2-4f(q) = D^2.\tag{4} $$

($1<q<10^8$ were checked).

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