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Peter asked here "Can a number be equal to the sum of the squares of its prime divisors?" and, it seems clear that if $$n=p_1^{a_1}\cdots p_k^{a_k},$$ and $$f(n):=p_1^2+\cdots+p_k^2$$ that then $n=f(n)$ only if $n$ is a square of a prime.

I almost proved that, but did not yet find so good and simple bounds to show that indeed it is possible only if $n$ is a square of a prime.

But, despite the fact I still do not have a proof, I decided to ask a little more general question:

If $$n=p_1^{a_1}\cdots p_k^{a_k},$$ is such that $$n=b_1p_1^2+...+b_kp_k^2$$ for some $b_1,...b_k \in \mathbb N$, then what is the density of the set of all such $n$ in $\mathbb N$?

I am also interested in how often do you find more than the one set $(b_1,...b_k)$ that solves $p_1^{a_1}...p_k^{a_k}=b_1p_1^2+...+b_kp_k^2$? That is, how often a function $n \to (b_1,...b_k)$ is multi-valued?

What is the biggest $n$ for which you did not find a solution?

Is the sequence of $n$´s for which there is no solution in OEIS?

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  • $\begingroup$ For small values of $n$, it shouldn't be hard for you to work out whether or not there is a solution. Then you can see for yourself whether that sequence is in the OEIS. Why not do that? $\endgroup$ – Gerry Myerson Jun 23 at 10:23
  • $\begingroup$ @GerryMyerson I am not skilled in programming, almost all calculations and proofs that I do are old-school fashioned, they are written on paper. Yes, I can see some small $n$´s that are not solutions. $\endgroup$ – Grešnik Jun 23 at 10:30
  • $\begingroup$ So, it's easy to see that if $n$ is prime there's no solution, right? That doesn't take any programming. If $n$ is the square of a prime, or any higher power of a prime, there is a solution, right? Now, $6=2\times3$, and there's no way to write $6$ as $4a+9b$. So at this point we know the sequence starts $2,3,5,6,7$ but not $8$ or $9$. It shouldn't be hard for you to do a few more cases without any programming, and then look up the results. Please, do it! $\endgroup$ – Gerry Myerson Jun 23 at 12:37
  • $\begingroup$ @GerryMyerson With only "by heart" calculations it seems to me that there are also no solutions for n=10,11,12,14,15,16..and so on. Yes, there are solutions for some prime n, why did you think there are not? $\endgroup$ – Grešnik Jun 23 at 12:53
  • $\begingroup$ How do you write $13$ as a linear combination of $13^2$? $\endgroup$ – Gerry Myerson Jun 23 at 12:55

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