# Find the number of natural numbers

$N$ is a natural number greater than 1 and less than 100. $F(1), F(2), \dots, F(n)$ are the factors of $N$ in such a way that $1=F(1)< F(2)< F(3)< \dots < F(n)=N$.

Further, $D= F(1)*F(2)+F(2)*F(3)+ \dots +F(n-1)*F(n)$. If $D$ is a factor of $N^2$, then how many values of $N$ will be there?

• How many prime numbers are there between $1$ and $100$ ? Does a prime number fulfill the conditions ? The hardest part is to prove that the prime numbers are the only ones. Maybe a proof by contradiction. – Fabien Jan 6 '15 at 10:49
If $N$ is prime, $D=N$, which divides into $N^2$
For intuition on the composites, if $N=pq$, with $p\lt q$ distinct primes, $D=p+pq+pq^2=p(1+q+q^2)$ which does not divide into $N^2$ as it exceeds $pq^2$ and does not have any factors of $q$.
To prove it for all composites, let $N$ be composite and $p$ be the smallest prime dividing $N$. $D \gt \frac {N^2}p$ as that is the last term in the sum for $D$. But $\frac {N^2}p$ is the largest proper factor of $N^2$. Thus if $D|N^2, D=N^2$