show that $k^4 - k^2 + 1 \neq n^2$ can never be a perfect square

Question

Let $\{x \}$ denote the fractional part of a number. E.g. $\{ \pi\} = 0.14159\dots$

In numerical computations I have that, for large values $k \gg 1$:

$$ \bigg\{ \sqrt{k^4 - k^2 + 1} \;\bigg\} \longrightarrow 0.5 $$

For example if $k = 300$ then the value inside the radical is: $8099730001$ and we have from a calculator:

$$ \sqrt{ 300^4 - 300^2 + 1} = 89998.499993 $$

That last decimal place might be wrong, but I get the sense this number is approaching a half-integer. Could it be that:

$$ \sqrt{k^4 - k^2 + 1} = k^2 \sqrt{ 1 - k^{-2} + k^{-4} } \approx k^2 \big( 1 - \frac{1}{2}k^{-2} \big) = k^2 - \frac{1}{2} $$

so I feel like this could never be an integer. How do I make that rigorous?

You should clarify that the curly brackets refer to the "fractional part" — Ben Grossmann, Nov 16 '16 at 01:14
A sufficient method is to calculate the limit of $\sqrt{k^4-k^2+1}-k^2$ as $k\to \infty$ — Ben Grossmann, Nov 16 '16 at 01:17

score 3 · Accepted Answer · answered Nov 16 '16 at 01:16

3

$(k^2 - 1)^2 = k^4 - 2 k^2 + 1 < k^4 - k^2 + 1 < (k^2)^2$ for all $k > 1$. There are no integers between $k^2-1$ and $k^2$.

answered Nov 16 '16 at 01:16

Robert Israel

1 Answers1