We'll get that $n =k^2+1$ for $k\in \mathbb N, k > 0$ $$\frac{n+11}{\sqrt{n-1}}=\frac{k^2+12}{k}$$. Then the rest is easy

I would like if u provide an answer @wardtnt1234 and also u should prove that $n=k^2+1$, its not that direct (ik it still easy but that would be like 4 points of deduction on the test)

Obviously $\sqrt{n-1}$ is rational, following which it becomes an integer since it is algebraic as well. Setting $\sqrt{n-1} = k$ for some positive integer $k$ we want $k \mid k^2+12$ which is equivalent to $k \mid 12$ following which we obtain $k = 1, 2, 3, 4, 6, 12$ which give values of $n$ as: \[n \in \{2, 5, 10, 17, 37, 145\}\]

starchan wrote: Obviously $\sqrt{n-1}$ is rational, following which it becomes an integer since it is algebraic as well. Setting $\sqrt{n-1} = k$ for some positive integer $k$ we want $k \mid k^2+12$ which is equivalent to $k \mid 12$ following which we obtain $k = 1, 2, 3, 4, 6, 12$ which give values of $n$ as: \[n \in \{2, 5, 10, 17, 37, 145\}\] is writing my solution in latex, but I just saw yours, and it is practically the same :u

Note that: $(\frac{n+11}{\sqrt{n-1}})^2$ is integer. $\implies \frac{n^2+22n+121}{n-1}$ is integer. $\implies n-1|n^2+22n+121$ $\implies n-1|144(n \ge 1)$ $\implies n-1={1,2,3,4,6,8,9,12,16,18,24,48,72,144}$ $\implies n={2,3,4,5,7,9,10,13,17,19,25,49,73,145}$ when replacing the integer condition we only have: $n=2,5,10,17,37,145$$\blacksquare$

I just saw and no one actually solved this it asks for the sum lol We need $\sqrt{n-1}=x$, for some integer $x$. Thus, we have $n=x^2+1$. Resubstituting, we get $$\frac{x^2+12}{x}\rightarrow x+\frac{12}{x}.$$We just need $x$ to be a divisor of $12$, which gives $1, 2, 3, 4, 6, 12$. The equivalent values of $n$ are $2, 5, 10, 17, 37, 145$, so our sum is $\boxed{216}$.

Let, $\sqrt{n-1}=m \implies n-1=m^2 \implies n=1+m^2.$ hence, $\frac{m^2+12}{m}$, needs to be an integer. Claim: $m$, is a divisor of $12.$ Proof: Note, that $m^2+12\equiv 12\pmod{m}$, in order for $12\pmod{m} \equiv 0\pmod{m}$, then, we need $m,$ to be a divisor of $12.$ Hence, $m=1,2,3,4,6,12.$ Now, we need to find $1^2+2^2+3^2+4^2+6^2+12^2+6=\boxed{216}.$ Nice one!

For some unknown reason i did not notice $n=k^2+1$ just kills the problem and instead i got a solution with a slightly annoying case checking. But here it is: Let $\frac{n+11}{ \sqrt{n-1}}=k$ for some positive integer $k$. $n+11=\sqrt{n-1}*k$ and squaring both sides we get $n^2+121+22n=(n-1)k^2 \implies n^2+n(22-k^2)+(121+k^2)=0$ and solving the quadratic we get $n=\frac{k^2-22+\sqrt{k^4-48k^2}}{2}$ or $n=\frac{k^2-22-\sqrt{k^4-48k^2}}{2}$ . It's obvious that $\sqrt{k^4-48k^2}$ is an integer so $k^4-48k^2$ is a perfect square therefore $k^4-48k^2=m^2$ for some positive integer $m$ but notice that $k^4-48k^2=m^2 \iff k^4-48k^2+24^2=m^2+24^2 \iff (k^2-24)^2=m^2+24^2$ which means $m^2+24^2$ is also a perfect square. Let $m^2+24^2=a^2 \implies (a-m)(a+m)=24^2$ and oh my, bashing this is a little tiring imo, i'll just give an example. Notice that $a-m<a+m$ so they can never be equal and $a+m$ must always have the bigger factors. We can have $a+m=288, a-m=2 \implies m=143 \implies k^2=\sqrt{143^2+24^2}+24$ This case gives $k^2=169 $ so $n=\frac{169-22+143}{2}=145$ or $n=\frac{169-22-143}{2}=2$. Doing the other cases you'll get the same solutions and the same answer like the comments above so i'm ending here.

Let $n = k^2 + 1$ (as $\sqrt{n-1}$ must be rational and therefore an integer, then $k \mid k^2 + 12 \iff k \mid 12$, giving the solutions easily. (In particular, $\sum_{d \mid 12} d^2 +1$)