# Find all natural numbers n such that…

Find all natural numbers N such that every natural divisor of N decreased by 1 is a square of some other natural number.
I've already found that 5, 37 and some other primes, such as first 5 Fermat's numbers meet the requirements, but the point is that N shouldn't be a prime number.
Also found out that the number should be a product of primes to the first power.

• What is a "natural" divisor? And explain the number 3? – gnasher729 Oct 31 '16 at 9:22
• Natural divisor D - a positive integer, such that N/D is also a positive integer – zzzizz Oct 31 '16 at 9:24
• DId you mean "every prime divisor..." ? – DonAntonio Oct 31 '16 at 9:24
• @zzzizz Then how come $\;3\;$ fulfills what you want? – DonAntonio Oct 31 '16 at 9:24
• No, every divisor, if there was a question about prime divisors it'd be a lot easier – zzzizz Oct 31 '16 at 9:25

I think one has to include $0$ as a natural number for this question, as any $N$ has the natural divisor $1,$ which decreased by $1$ gives $0,$ which is the square of $0$ (but not the square of a positive nnatural (some say only positive integers are natural numbers). Let's put that triviality aside-- Now you're looking for non-prime examples, so consider $N=10,$ whose (non-1) divisors are $2,5,10$ and these decreased by $1$ are all squares, $1,4,9.$
added: I think it's not hard to show that one cannot have other $N$ than $10$ which are even and divisible by some other odd prime $p$, by considering only the facts that one would need both $p=x^2+1$ and $2p=y^2+1,$ among whatever other equations if there are more prime divisors. One gets to $y=x+1$ and keeps going. So what's left is the case of odd $N.$ [That's if the last paragraph works out for a proof of that case.]