Dividing sequences

The nth term for the first sequence is $(\frac{n(n+1)}{2})^{2}$ and the nth term of the second sequence is $\frac{n(n+1)(2n+1)}{6}$ .

After dividing, the answer simplifies to $\frac{3n(n+1)}{2(2n+1)}$ . This has to be an integer.

$gcd(2n, 2n+1)=1$ so $gcd(n, 2n+1)=1$

Also, $gcd(2n+2, 2n+1)=1$ so $gcd(n+1, 2n+1)=1$

Therefore, for $\frac{3n(n+1)}{2(2n+1)}$ to be an integer, $2n+1|3$ .

So $n=1$ .

Looking back at the sequences, we see that when $n=1$ , we get $\frac{1}{1}=1$ which is an integer, so there is one value for n.