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Oh Good one! Thanks!
By trying all of 2,3,5,7,11,13,17,19, and 23?
It is easily checked that 677 is prime.
PS: In general:
In the above problem:
Isn't it just the a^2 - b^2 formula?
There are things called aurifeuillian factorizations.
This one could be the basis for many others. But like Aurifeuille who used it for n = 14 in 1871 there is much trial and error.
How did you come into that formula?
What is the largest prime factor of 5^8 + 2^2?
As bobbym pointed out, n must be a perfect square. n=1 is one possibility. For the others, it can be easily checked that all odd perfect squares greater than 1 and less than 1000 are have at most two distinct prime factors in their factorization. Thus the possibilities for n>1 are:
where p and q are distinct primes and a, b positive integers.
The number of positive divisors of n are – i.e. there are positive divisors. So the possibilites are and . (Not ; that would make n too large.)
There are only two such possible, namely and . The number of positive divisors for each number is 9, which does divide each number.
Therefore the answer to your question is: There are 5 odd numbers less than 1000 which are divisible by their number of positive divisors, namely 1, 9, 225, 441, and 625.
That is how I did it. You just square 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31 and check.
Ooh, do we search them manually?
3^2 = 9
And why not 9 as bob told?
I'm not following this thread at all.
I am not sure what you are exactly asking so I will answer every possible question.
Odd numbers have odd divisors.