Let $n^2+32n+8=k^2$ Then, $(n+16)^2-k^2=248=2^3\cdot 31$ Then, the only two pairs of factors that work are $(2, 124)$ and $(4, 62)$, $\implies n=47, 17$

$k^2–n^2=8(4n+1)$ case work@wolfusa below Your one is good but very lazy

$(n+16)^2>n^2+32n+8>(n+5)^2$ so you just have to solve $n^2+32n+8=(n+6)^2$ or $n^2+32n+8=(n+7)^2$ or... or $n^2+32n+8=(n+15)^2$

I don't think it's a good idea to bump old threads, but I have a very curious question. Why is the Hong Kong TST so easy compared to every other China competition? China TST is exponentially harder.

Maybe because there are much fewer people in HongKong than in China?

YanYau wrote: Find all positive integer(s) $n$ such that $n^2+32n+8$ is a perfect square. For $n\ge109$ we have $$(n+15)^2<n^2+32n+8<(n+16)^2$$so it suffices to check for $n\le108$. For $47<n\le108$ we have $$(n+14)^2<n^2+32n+8<(n+15)^2$$so it suffices to check for $n\le47$. If $n=\boxed{47}$ then $3721=61^2$ is a perfect square. For $27<n<47$ we have $$(n+13)^2<n^2+32n+8<(n+14)^2$$so it suffices to check for $n\le27$. For $17<n\le27$ we have $$(n+12)^2<n^2+32n+8<(n+13)^2$$so it suffices to check for $n\le17$. If $n=1$, $41$ is not a perfect square. If $n=2$, $76$ is not a perfect square. If $n=3$, $113$ is not a perfect square. If $n=4$, $152$ is not a perfect square. If $n=5$, $193$ is not a perfect square. If $n=6$, $236$ is not a perfect square. If $n=7$, $281$ is not a perfect square. If $n=8$, $328$ is not a perfect square. If $n=9$, $377$ is not a perfect square. If $n=10$, $428$ is not a perfect square. If $n=11$, $481$ is not a perfect square. If $n=12$, $536$ is not a perfect square. If $n=13$, $593$ is not a perfect square. If $n=14$, $652$ is not a perfect square. If $n=15$, $713$ is not a perfect square. If $n=16$, $776$ is not a perfect square. If $n=\boxed{17}$, $841=29^2$ is a perfect square.