broniran wrote:

But $n=1$ works!

Well, yes. People make mistakes. Care to complete the solution?

jmerry wrote: Well, yes. People make mistakes. Care to complete the solution? Sure. Let $n^4-4n^3+22n^2-36n+18=k^2$. So $(n^2 - 2n + 9)^2-k^2=63$. Factor this $(n^2-2n+9-k)(n^2-2n+9+k)=63$. Now use factors of $63$ and do casework to complete. Not sure what broniran was trying to do though, maybe someone else can explain his/her solution?

Fang-jh wrote: Find all positive integers $ n$ such that $ n^4-4n^3+22n^2-36n+18$ is a perfect square. You can solve this in integers. If $n\in(-\infty,-4]\cup[6,+\infty)$, then $(n^2 - 2n + 9)^2 > n^4 - 4n^3 + 22n^2 - 36n + 18 > (n^2 - 2n + 8)^2$ (the first inequality is trivial; the second becomes quadratic after a simplification), so no solutions. It's only left to check $n\in\{-3,\ldots,5\}$, which only gives $n=-1,1,3$.

gethd wrote: Fang-jh wrote: Find all positive integers $ n$ such that $ n^4-4n^3+22n^2-36n+18$ is a perfect square. You can solve this in integers. If $n\in(-\infty,-4]\cup[6,+\infty)$, then $(n^2 - 2n + 9)^2 > n^4 - 4n^3 + 22n^2 - 36n + 18 > (n^2 - 2n + 8)^2$ (the first inequality is trivial; the second becomes quadratic after a simplification), so no solutions. It's only left to check $n\in\{-3,\ldots,5\}$, which only gives $n=-1,1,3$. Indeed and we exclude $-1$ since they wanted positive integers edit: oops completely skimmed over that part , thanks.

gethd wrote: You can solve this in integers. ...is what he emphasized.

Obvious solution is $n=1$. Let $n^4-4n^3+22n^2-36n+18=m^2$ or $n^4-4n^3+4n^2+18n^2-36n+18=m^2$. This is equal to $(n^2-2n)^2+18(n^2-2n)+18=m^2$ or $(n^2-2n)^2+18(n^2-2n)+81-63=m^2$ and $(n^2-2n+9)^2-63=m^2$. Finally $(n^2-2n+9-m)(n^2-2n+9+m)=63$ We have few options: $(n^2-2n+9+m;n^2-2n+9-m)=(1,63);(3,21);(7,9)$. This has natural solutions for $(3,21)$. Solve the equation $n^2-2n+9=12$ and $n=3$ or $n=-1$. Finally, solutions are $n=1;3$