How could one prove that there are no positive integers for which a given expression is a perfect square?

Question

How can one find all natural $n$ for which $6n-2n^2+32n^4+25$ is a perfect square? In other words, for which $n\in \mathbb{N}$ is $6n-2n^2+32n^4+25 = k^2$ where $k\in \mathbb{N}$. I am fairly certain that there are no positive integer solutions, however, I struggle in finding a proof for this. If there was a general approach for polynomial expressions, an explanation for finding whether it can equal a perfect square would be welcome. I know of approaches using the fact that $k^2\equiv0,1\pmod 3$ and $k^2\equiv0,1\pmod 4$, however, I can't find a way to use this in this particular example.

Answer 1

From where you left off:

$$32n^2 - 2n^2 = k^2 - 6n - 25$$

Completing the square on the $n^2$ term in LHS:

$$32(n^2 - \frac{1}{32})^2 - \frac{1}{32} = k^2 - 6n - 25$$

$$\big(\frac{k}{\sqrt{2}}\big)^2 = (4(n^2 - \frac{1}{32}))^2 + 3n + \frac{25}{2} - \frac{1}{32}$$

$$\big(\frac{k}{\sqrt{2}}\big)^2 = (4(n^2 - \frac{1}{32}))^2 + 3n + \frac{799}{64}$$

$\require{cancel}\xcancel{\text{We have arrived in the form } c^2 = a^2 + b^2. \text{We obtain, } n = \frac{b^2}{3} - \frac{799}{192} = \frac{64b^2 - 799}{192}. \text{ If } n \text{ is a positive integer, then }192 \text{ } | \text{ } (64b^2 - 799). \text{ However, there is contradiction that }64b^2 - 799 \text{ is odd, but }192 \text{ even}. \text{ Hence, }n \text{ cannot be integral.}}$

$$\big(\frac{k}{\sqrt{2}}\big)^2 = \big(4n^2 - \frac{1}{8}\big)^2 + \frac{192n + 799}{64}$$

$$\big(\frac{k}{\sqrt{2}}\big)^2 = \frac{\big(32n^2 - 1\big)^2 + 192n + 799}{64}$$

If $k$ is even, then $\frac{k^2}{2} \in \mathbb{N}$.

Hence, $64 \text{ | } \big(32n^2 - 1\big)^2 + 799 + 192n$ for $n \in \mathbb{N}$.

Assume $64 \text{ | } \big(32n^2 - 1\big)^2 + 799$

$$\big(32n^2 - 1\big)^2 + 799 = 1024n^4 - 64n^2 + 800$$

Clearly, $64 \text{ | } 1024n^4$ and $64 \text{ | } -64n^2$ but $800$ is not divisble by $64$, hence we have a contradiction, $\implies$ $n$ is not an integer.

$\require{cancel} \xcancel{ \text{ If k is odd }, k^2 \text{ is odd.} \text{ Then, back to the beginning: }\frac{1}{2} k^2 = \frac{1}{2}\big(32n^4 - 2n^2 + 6n + 25 \big) k^2 = 32n^4 - 2n^2 + 6n + 5^2 \implies k^2 = a^2 + b^2 n \in \mathbb{N} \implies 32n^4 - 2n^2 + 6n \in \mathbb{N}. \text{ Furthermore, } k \in \mathbb{N} \text{ restricts to the only triple possible: }\{k, a, b\}: \{13, 12, 5\}. \text{ Let } f(n) = 32n^4 - 2n^2 + 6n. \text{ Hence, we must have }f(n) = 144. \text{ However, } f(1) = 36, f(2) = 516\text{, indicating solution of } f(n) = 144 \text{ exists for n } \in (1, 2), \text{ and contradicting that n is integral.}}$

EDIT:

I cannot believe I did not think of this beforehand:

For $6n−2n^2+32n^4+25$ to be a perfect square, it can be least expressed in the form $(an^2 + bn + c)^2$ for some $a, b, c$ coeff.

Expanding, $(an^2 + bn + c)^2$, gives $$(a^2)n^4 + (2ab)n^3 + (2ac + b^2)n^2 + (2bc)n + c^2$$

Solving simultaneously, we first have:

$$\begin{cases} a^2 = 32, a = \pm\sqrt{32} \\ c^2 = 25, c = \pm5 \\ 2bc = 6, b = \frac{\pm3}{\sqrt{32}} \end{cases}$$

Since there is no $n^3$ term, $2ab = 0$. However, both $a, b ≠ 0$ and we have a contradiction. Furthermore, the irrational coefficients $a, b$ suggest that $an^2 + bn + c \notin \mathbb{N}, n \in \mathbb{N}$.