integer $n$ for which $n^6+3n^5-5n^4-15n^3+4n^2+12n+3$ is a perfect Square Prove that the no integer $n\;,$ for which $n^6+3n^5-5n^4-15n^3+4n^2+12n+3$ is a perfect 
Square.
My Try:: We can write $(n^6+3n^5-5n^4-15n^3+4n^2+12n+3) = (n^3+an^2+bn+c)^2$
Now Here we have to find values of $a,b,c$.
But 
this become very complex.
is there is any other way to solve the above question.
Help me
Thanks

Modulo $4$ we have
$$\eqalign{n^6+3n^5-5n^4-15n^3+4n^2+12n+3
  &\equiv n^6-n^5-n^4+n^3+3\cr
  &\equiv n^3(n+1)(n-1)^2+3\cr
  &\equiv3\ ;\cr}$$
but a square can only be congruent to $0$ or $1$ modulo $4$.
Reason for the last step: if $n$ is even then $n^3$ is a multiple of $4$, while if $n$ is odd then $(n-1)^2$ is a multiple of $4$.
