14. Determine the smallest integer $n$ with $n \geq 2$ such that
$$
\sqrt{\frac{(n+1)(2 n+1)}{6}}
$$
is an integer.
14. Answer: 337
Solution. Assume that
$$
\sqrt{\frac{(n+1)(2 n+1)}{6}}=m
$$
and so
$$
(n+1)(2 n+1)=6 m^{2} .
$$

Thus $6(n+1)(2 n+1)$, implying that $n \equiv 1$ or $5(\bmod 6)$.
Case 1: $n=6 k+5$.
Then $m^{2}=(k+1)(12 k+11)$. Since $(k+1)$ and $(12 k+11)$ are relatively prime, both must be squares. So there are positive integers $a$ and $b$ such that $k+1=a^{2}$ and $12 k+11=b^{2}$. Thus $12 a^{2}=b^{2}+1$. But, as $12 a^{2} \equiv 0(\bmod 4)$ and $b^{2}+1 \equiv 1$ or $2(\bmod 4)$, there are no integers $a$ and $b$ such that $12 a^{2}=b^{2}+1$. Hence Case 1 cannot happen.
Case 2: $n=6 k+1$.
Then $m^{2}=(3 k+1)(4 k+1)$. Since $3 k+1$ and $4 k+1$ are relatively prime, both must be squares, So there are positive integers $a$ and $b$ such that $3 k+1=a^{2}$ and $4 k+1=b^{2}$. Then $3 b^{2}=(2 a-1)(2 a+1)$. Observe that in the left-hand side, every prime factor except 3 has an even power. So neither $2 a-1$ nor $2 a+1$ can be a prime other than 3 .
Now we consider positive integers $a$ such that neither $2 a-1$ nor $2 a+1$ can be a prime other than 3. If $a=1$, then $b=1$ and $n=1$. So we consider $a \geq 2$. The next smallest suitable value for $a$ is 13 . When $a=13$, we have
$$
3 b^{2}=25 \times 27
$$
and so $b=15$, implying that $k=56$ and so $n=6 k+1=337$.