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When is $3136 - 6272 n + 3136 n^2 + n^{14} =z^2$ for $z \in \mathbb N$?

Here was my approach: Consider $3136 - 6272 n + 3136 n^2 + n^{14} \pmod 5$. Since any integral square must be equivalent to 0, 1, or 4, modulo 5, we can rule out $n=5k+3$, since, making the substitution, we get

$$3136 - 6272 n + 3136 n^2 + n^{14} = 12544 + 62720 k + 78400 k^2 + (3 + 5 k)^{14}$$

which is always equivalent to 3 mod 5. Therefore $x \neq 5k+3$.

Continuing, we know a square must be equivalent to 0, 1, 3, or 4, modulo 6. However, the polynomial is equivalent to 2 and 5 modulo 6 whenever $n$ is equivalent to 2 or 5 modulo 6 respectively.

Hence, $n \neq 5k + 3, n \neq 6k+1, n \neq 6k+4, \cdots$

I suppose we can eliminate many arithmetic progressions from the possibilities of $n$, but that doesn't solve the case of whether the polynomial evaluated at some large prime for example might give output to a square integer. I'm not exactly sure how else to proceed here - any thoughts?

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  • $\begingroup$ Just realized, the substitution $1-u=n$ gives $3136u^2 + (u-1)^{14}$. $\endgroup$
    – Snared
    Commented Nov 23, 2023 at 5:22
  • $\begingroup$ I just realized 3136 is a square. Maybe something with a difference of squares? $(z+56n-56)(z-56n+56)=n^{14}$ $\endgroup$
    – Mike
    Commented Nov 23, 2023 at 5:31
  • $\begingroup$ @Mike using $z \mapsto n^7$ gives $(z+56n-56)(z-56n+56) = n^{14} - O(n^2)$. I wonder if it's possible to determine whether or not there is some mapping of $z$ to $O(n^7)$ which can keep the $n^{14}$ and zero out the $O(n^2)$.. $\endgroup$
    – Snared
    Commented Nov 23, 2023 at 5:48
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    $\begingroup$ Easy bound, your number is always bigger than $(n^7)^2$ but it is smaller than $(n^7 + 1)^2 $ unless $n$ is quite small $\endgroup$
    – Will Jagy
    Commented Nov 23, 2023 at 5:48
  • $\begingroup$ @WillJagy not sure I quite understood your comment, which number are you referring to? Something like, if $z^2 = f(n)$ then $(n^7)^2 < z^2 < (n^7+1)^2$? $\endgroup$
    – Snared
    Commented Nov 23, 2023 at 5:54

1 Answer 1

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It may be checked that the expression

$$3136 - 6272n + 3136n^2 + n^{14} > (n^7+1)^2$$

reduces to

$2 \le n \le 3$. This can be seen by analyzing derivatives and critical points of both functions numerically. Then, since $3136 - 6272n + 3136n^2 > 0$ for $n>0$, we have that $n^{14} < 3136 - 6272n + 3136n^2 + n^{14}$.

It follows that for $n>3$,

$$u^2 < 3136 - 6272n + 3136n^2 + n^{14} < (u+1)^2$$

our expression lies strictly between two consecutive perfect squares, and is thus not a perfect square herself. Here $u=n^7$.

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  • $\begingroup$ (Thanks for the solution Will Jagy in the comments) $\endgroup$
    – Snared
    Commented Nov 23, 2023 at 21:08

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