in order to be a product of $3$ or more consecutive integers we must have $3|n^4 + 8n + 11$, which is trivially impossible, (by simple mod checking). so we have $a.(a+1)=n^4 + 8n + 11\implies (2a+1)^2=4.n^4+32n+45$, but $(2n^2)^2<4.n^4+32n+45<((2n^2)+1)^2$ if, $n\geq 10$, so checking values for $n\leq 9$ gives, $n=1$.

If this is a product of 3 integers then this needs to be divisible by $3$ which is impossible, so $n \le 10$, the only value that works is if $n=1.$

sayanjoddar wrote: in order to be a product of $3$ or more consecutive integers we must have $3|n^4 + 8n + 11$, which is trivially impossible, (by simple mod checking). so we have $a.(a+1)=n^4 + 8n + 11\implies (2a+1)^2=4.n^4+32n+45$, but $(2n^2)^2<4.n^4+32n+45<((2n^2)+1)^2$ if, $n\geq 10$, so checking values for $n\leq 9$ gives, $n=1$. why do u consider $n\geq 10$ and how do you check the values?

Addendum: I checked symbolab, and the intervals made sense, but how did you get the intuition to demarcate the line between n>= 10 and n<=9, as well as n=1?

RenheMiResembleRice wrote: Addendum: I checked symbolab, and the intervals made sense, but how did you get the intuition to demarcate the line between n>= 10 and n<=9, as well as n=1? If $n>9$ as shown by sayanjoddar is bounded between two consecutive perfect squares so it cannot be a perfect square, then $n<10$, and doing cases manually you get $n=1$