<s>[INST] Can't solve a quartic equation I'm trying to solve an algebraic question.The question wants me to solve $n^4+2n^3+6n^2+12n+25=m^2$.The question also states that n is a positive integer and the answer for $n^4+2n^3+6n^2+12n+25$ is a square number. Here's how I tried to solve it:
$$n^4+2n^3+6n^2+12n+25=\\n^4+6n^2+2n^3+12n+25=\\
n^2(n^2+6)+2n(n^2+6)+5^2=\\
(n\sqrt {n^2+6})^2+2n(n^2+6)+5^2.\\$$
Because $a^2+2ab+b^2=(a+b)^2$,so
$\sqrt {(n\sqrt {n^2+6})^2}\cdot\sqrt5^2=n(n^2+6)$
Then:
$$n\sqrt {n^2+6}\cdot 5=n(n^2+6)\\
\sqrt {n^2+6}\cdot 5=n^2+6\\
25(n^2+6)=n^4+12n^2+36\\
n^4+12n^2+36=25n^2+150\\
n^4-13n^2-114=0\\
(n^2+6)(n^2-19)=0\\
n^2=19\\
n=\sqrt 19$$
But $n$ is a positive integer.
Can anyone help?
 [/INST] $$ ( n^2 + n + 2 )^2 = n^4 + 2n^3 + 5n^2 + 4n + 4 $$
$$ ( n^2 + n + 3 )^2 = n^4 + 2n^3 + 7n^2 + 6n + 9$$ 
The second one is larger than yours, meaning yours cannot be square, when
$$ 7n^2 + 6n+9 - 6n^2 - 12 n - 25 > 0 \; , \;  $$
$$ n^2 - 6n - 16 > 0 \; , \;  $$
$$ (n-8)(n+2) > 0 \; .    $$
You need check only $0 \leq n \leq 8.$ 
ADDED: the quartic in the question also lies strictly between consecutive squares when $n \leq -3.$ The squares are just $n=-2,0,8.$
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