Let $M=2n^{3k} + 4n^k + 10$. If it is a product of any three consecutive integers, then it must be a multiple of $3$ because one of these three integers must be a multiple of $3$. However, we claim that $M\equiv 1\pmod{3}$ which means it is not a multiple of $3$. To show this, let $m=n^k$, then $M= 2m^3 + 4m+10$.

• If $m\equiv 0\pmod{3}$, then $N\equiv 0 + 0 + 10\equiv 1\pmod{3}$
• If $m\equiv 1\pmod{3}$, then $N\equiv 2 + 4 + 10\equiv 1\pmod{3}$
• If $m\equiv -1\pmod{3}$, then $N\equiv -2 -4 + 10\equiv 1\pmod{3}$

Meanwhile, we can show $M$ cannot be a product of two consecutive integers either. In order to show this, let's exam all the possible cases below where $p$ is an integer.

• $3p(3p+1) \equiv  0\pmod{3}$
• $(3p+1)(3p+2)\equiv 2\pmod{3}$
• $(3p+2)(3p+3)\equiv 0\pmod{3}$

But we know $M\equiv 1\pmod{3}$. Therefore $M$ cannot be a product of two consecutive integers.

Hence, we conclude the claim holds.