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I'm doing a Number Theory question, and if someone could offer a hint, that would be greatly appreciated;

The question is: Find the sum of the perfect square divisors of the smallest integer with exactly 6 perfect square divisors.

My reasonings:

  • My method so far has been simply listing out pairs of possible exponents that are multiples of 2, and listing them out, or using combinations to calculate the total number of combinations. Obviously, this brute-force method is not working out well, and if anyone could provide a hint, that would be greatly appreciated.
  • I know how to find the product of all divisors, but not their sum. Furthermore, how would one "target" only perfect square divisors? Perhaps complementary counting?
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    $\begingroup$ the smallest number with exactly six square divisors is going to be a square itself. Try the first several squares, see what happens $\endgroup$ – Will Jagy Dec 29 '17 at 17:09
  • $\begingroup$ Do you know how counting the number of divisors of a number given its prime decomposition works? It's not too difficult to adjust to square divisors. $\endgroup$ – Arthur Dec 29 '17 at 17:10
  • $\begingroup$ Of course, because the only possible exponents can be 1, and 2, which would result in a square. $\endgroup$ – DarkRunner Dec 29 '17 at 17:10
  • $\begingroup$ @DarkRunner the exponents can also be 4, 6,... $\endgroup$ – idok Dec 29 '17 at 17:11
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    $\begingroup$ You can have more than one prime factor! if $n=\prod p_i^{a_i}$ then the number of square divisors of $n$ is $\prod \left(1+\lfloor \frac {a_i}2\rfloor\right)$. $\endgroup$ – lulu Dec 29 '17 at 17:16
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Conversation in the comments is correct, but I would suggest the following approach. First we find the lowest number with exactly $6$ divisors, and then we square that number. This squared number will then have exactly $6$ square divisors.

To find the number, we use the number of divisors formula $$d\left(\prod p_i^{a_i}\right) = \prod(a_i +1)$$ Now the divisors of $6$ are $6,1$ and $2,3$, so now we choose the lowest primes and attach these exponents to them as follows. $$2^5=32 \quad 2^2*3=12$$ Now we have the lowest number with exactly $6$ divisors, $12$. So the lowest number with 6 square divisors is $144$. The square divisors are the squares of the divisors of $12$, so now we add those up to get our answer. $$1^2+2^2+3^2+4^2+6^2+12^2=210$$

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  • $\begingroup$ Problem asks for sum, not product, of square divisors. $\endgroup$ – Gerry Myerson Dec 29 '17 at 17:50
  • $\begingroup$ Oh, sorry. I'll fix that. $\endgroup$ – Isaac Browne Dec 29 '17 at 17:59
  • $\begingroup$ Wow; so essentially working backwards towards the solution. $\endgroup$ – DarkRunner Dec 30 '17 at 3:13
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In order to have exactly $6$ square divisors, a number $n$ is either of the form $n=p^{10}m$ with $p$ a prime and $m$ square-free or of the form $n=p^4q^2m$ with $p$ and $q$ distinct primes and $m$ square-free. In the first case the square divisors are $1+p^2+p^4+p^6+p^8+p^{10}$ and the smallest possibility is $n=2^{10}=1024$. In the second case the sum of the square divisors is $1+p^2+q^2+p^4+p^2q^2+p^4q^2$ and the smallest possibility has is $n=2^4\cdot3^2=144$. So $n=144$ is the smallest possibility, and the sum of its square divisors is $1+4+9+16+36+144=210$.

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