Consider $a=\frac{b!}{4}-1$. Then, $a^2+b!$, $(a+2)^2+1!$, $(a+2)^2+2!$ works.

Not a solution but related: By $\mod 8$ and finite case check for small values, you can prove that at maximum $4$ consecutive numbers are emphatic. I don't know if there are infinitely $n$ with $n,n+1,n+2,n+3$ emphatic, but $n=168$ works.

It suffices to show that there are infinitely many solutions to $a^2 + b! = m^2$, as $1 = 1!$ and $2! = 2$. We want $(m-a)(m+a) = b!$, so take $b\geq 4$ and we can have $m-a = 2$, $m+a = \frac{b!}{2}$, i.e. $a = \frac{b!}{4} - 1$, $m = \frac{b!}{4} + 1$.

IAmTheHazard wrote: Call a positive integer emphatic if it can be written in the form $a^2+b!$, where $a$ and $b$ are positive integers. Prove that there are infinitely many positive integers $n$ such that $n$, $n+1$, and $n+2$ are all emphatic. Allen Wang $x^2+1=a^2+1!$,$x^2+2=a^2+2!$ So we need to prove that there is infinite solution to the equation $x^2+3=a^2+b!$ For $x=a+c$ we get $c(c+2a)=b!-3$ And for $b>=3$ we can get $c=3$ and we are done.

Note that $a^2 + 1, a^2 + 2$ are clearly emphatic for any positive integer $a$. Claim: There are infinitely many perfect square that are emphatic. Proof: For any $b > 2$, we note that if $b! = 4k$, then $(k+1)^2 = (k-1)^2 + b!$, so $(k+1)^2$ is emphatic. There are clearly infinitely many possible values of $k$, so the claim is proven. $\square$ Now, there exist infinitely many $a$ where $a^2$ is emphatic, and $a^2 + 1, a^2 + 2$ are also emphatic, so taking $n = a$ for any such $a$ works.

Let $d$ be a positive divisor of $\frac{m!}{4}$ where $m\geq 4$ is an integer. Let $n=(\frac{m!}{4d}+d)^2$. Then the following equalities hold. $$n=(\frac{m!}{4d}-d)^2+m!$$ $$n+1=(\frac{m!}{4d}+d)^2+1!$$ $$n+2=(\frac{m!}{4d}+d)^2+2!$$ Therefore $n$ is emphatic. Changing values of $m$ and $d$ gives us infinitely many emphatic positive integers. Therefore there exists infinitely many emphatic positive integers.

Very Easy. Take $b$ such $2|b$ and $a=\frac{(b-1)!-b}{2}$ And take $n=a^2+b!$. Note that $n$ is a square number. $n+1=n+1!$ and $n+2=n+2!$ so there exists infinite triplets $(n,n+1,n+2)$ $\square$