Can't solve a quartic equation

Question

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?

Answer 1

$$ ( n^2 + n + 2 )^2 = n^4 + 2n^3 + 5n^2 + 4n + 4 $$

$$ ( n^2 + n + 3 )^2 = n^4 + 2n^3 + 7n^2 + 6n + 9$$ 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 $n=-2,0,8.$

Answer 2

I believe (correct me if I'm wrong) that $a^2-b^2$ only has factors $1, a^2-b^2, (a-b), (a+b)$ if $a$ and $b$ are coprime. Using this, we have that:

$$n^4+2n^3+6n^2+12n+25=m^2 \to n(n^3+2n^2+6n+12)=m^2-25$$

If we assume that $m\neq 5k,k\in\Bbb Z$, by what I stated earlier we have that either $n=1$ and $(n^3+...)=m^2-25$, which leads to $m=\pm\sqrt{46}$, or $(n^3+...)=1$ and $n=m^2-25$, which leads to $n\approx -1.896\to m\approx\sqrt{23.104}$, or that $(n^3+...)=n\pm 10\because m^2-25=(m+5)(m-5)$ and these factors are $10$ apart. Neither of these yields integer solutions, so we can safely discard them too.

In short, this is impossible unless $m$ is a multiple of $5$. Now see if you can find if it works when $m$ is.

Answer 3

Above equation shown below:

$n^4+2n^3+6n^2+12n+25=m^2$

As pointed out by Will Jagy the only positive integer solution to above equation is $(n,m)=(8,75)$

Also $m$ is a multiple of five as mentioned by Rhyes hughes