Let $n$ be an odd integer. Is $$(1+c^2)^n-\lfloor(1+c^2)^{n/2}\rfloor^2<(1+c^2)^{(n+1)/2}$$ true for all integers $c>1$?
Notes:
$c=1$ has a counterexample $2^{31}-\lfloor2^{31/2}\rfloor^2>2^{16}$; there are probably infinitely many of these. No counterexamples have been found for $(c,(n-1)/2)\in[2,20]\times[1,50]$.
The bounds $x-1\le\lfloor x\rfloor\le x+1$ are too weak to prove anything non-trivial. This is because we are looking for a Diophantine approximation of half-integer powers of $1+c^2$ to an integer, and this doesn't seem straightforward at all.
This inequality comes from an attempt to bound the difference $|b-a|$ when an integer can be represented as $a^2+b^2$ in more than one way. It turns out that this is a very difficult thing to do, as we need to know and there is only a paucity of literature on the irrationality measure involving inverse trigonometric functions.
I'd argue that it is still possible to do via $\lfloor x\rfloor\geq x-1$: $$\left(1+c^{2}\right)^{n}-\lfloor\left(1+c^{2}\right)^{\frac{n}{2}}\rfloor^{2}\leq\left(1+c^{2}\right)^{n}-\left(\left(1+c^{2}\right)^{\frac{n}{2}}-1\right)^{2}$$ Expand the RHS to get: $$\left(1+c^{2}\right)^{n}-\left(\left(1+c^{2}\right)^{\frac{n}{2}}-1\right)^{2}=2\left(1+c^{2}\right)^{\frac{n}{2}}-1<2\left(1+c^{2}\right)^{\frac{n}{2}}$$ And since $c\geq 2$, $\sqrt{c^2+1}\geq\sqrt5>2$: $$2\left(1+c^{2}\right)^{\frac{n}{2}}<\sqrt{c^2+1}\left(1+c^{2}\right)^{\frac{n}{2}}=2\left(1+c^{2}\right)^{\frac{n+1}{2}}$$ Unless, I am not understanding the problem correctly. I don't think the part mentioning that $c$ is odd was even nessacary for this argument. I did check this graphically and there are counterexamples, only for $c=1$ though, as only for $c=1$ is $\sqrt{c^2+1}<2$