IMPORTANT: I posted this as I thought I found a generalized result, which turned out to be false (by a trivial flaw), but a nice lemma still remains (however, this can be closed now)
Lemma:
If $p\equiv 1\pmod{8}$, then $$\bigg(\frac{(\frac{p-1}{4})!}{p}\bigg)=-1$$
This problem below appeared in a USA contest (I think) some time ago:
Let $p$ be a prime number and $k\in\mathbb{N}$ such that $p=8k+1$. Suppose that $$\binom{4k}{k}\equiv r\pmod{p}$$ Prove that $\sqrt{r}$ is not an integer.
Now this is actually true (By the way, try the problem yourself, it is very beautiful). So under this assumptioin, lets prove the lemma
Proof:
When $v_2(p-1)=3$
This gives us $p\equiv 9\pmod{16}\Rightarrow p\equiv 1\pmod{8}$. Lets write $p=8k+1$. So by the above problem i just stated, we get:
$$\bigg(\frac{\binom{4k}{k}}{p}\bigg)=-1$$
Lets start interpreting this:
$$\bigg(\frac{\binom{4k}{k}}{p}\bigg)=-1\Leftrightarrow \bigg(\frac{(4k)!}{p}\bigg)=-\bigg(\frac{\frac{1}{k!\cdot(3k)!}}{p}\bigg)$$
Notice that
$$1=\bigg(\frac{1}{p}\bigg)=\bigg(\frac{\frac{1}{k!\cdot(3k)!}}{p}\bigg)\cdot\bigg(\frac{k!\cdot(3k)!}{p}\bigg)$$
so $$\bigg(\frac{k!\cdot(3k)!}{p}\bigg)=\frac{1}{\bigg(\frac{\frac{1}{k!\cdot(3k)!}}{p}\bigg)}=\bigg(\frac{\frac{1}{k!\cdot(3k)!}}{p}\bigg)$$
Thus, substituting in
$$\bigg(\frac{(4k)!}{p}\bigg)=-\bigg(\frac{\frac{1}{k!\cdot(3k)!}}{p}\bigg)$$
we have
$$\bigg(\frac{(4k)!}{p}\bigg)=-\bigg(\frac{k!\cdot(3k)!}{p}\bigg)\Leftrightarrow\bigg(\frac{(3k)!}{p}\bigg)\cdot\bigg(\frac{(3k+1)(3k+2)...4k}{p}\bigg)=-\bigg(\frac{(3k)!}{p}\bigg)\cdot\bigg(\frac{k!}{p}\bigg)$$
so we get
$$\bigg(\frac{(3k+1)(3k+2)...4k}{p}\bigg)=-\bigg(\frac{k!}{p}\bigg)$$
which is equivalent to
$$\big((3k+1)(3k+2)...4k\big)^{\frac{p-1}{2}}\equiv-\big(k!\big)^{\frac{p-1}{2}}\pmod{p}$$
Multiplying both sides by $2^{\frac{p-1}{2}}$, we have:
$$\big((6k+2)(6k+4)...8k\big)^{\frac{p-1}{2}}\equiv-\big(2\cdot4\cdot6...\cdot2k\big)^{\frac{p-1}{2}}\pmod{p}$$
But
$$\big((6k+2)(6k+4)...8k\big)^{\frac{p-1}{2}}\equiv\big(-(2k-1)\cdot-(2k-3)...\cdot-1\big)^{\frac{p-1}{2}}\equiv \big( (-1)^k\cdot1\cdot3...\cdot(2k-1)\big)^{\frac{p-1}{2}}\pmod{p}$$
$\frac{p-1}{2}=4k$, so
$$\big( (-1)^k\cdot1\cdot3...\cdot(2k-1)\big)^{\frac{p-1}{2}}\equiv\big(1\cdot3...\cdot(2k-1)\big)^{\frac{p-1}{2}}\pmod{p}$$
Substituting in
$$\big((6k+2)(6k+4)...8k\big)^{\frac{p-1}{2}}\equiv-\big(2\cdot4\cdot6...\cdot2k\big)^{\frac{p-1}{2}}\pmod{p}$$
we get the following:
$$\big(1\cdot3...\cdot(2k-1)\big)^{\frac{p-1}{2}}\equiv-\big(2\cdot4\cdot6...\cdot2k\big)^{\frac{p-1}{2}}\pmod{p}$$
Finally, by multiplying both sides by $\big(2\cdot4\cdot6...\cdot2k\big)^{\frac{p-1}{2}}$, we get:
$$(2k)!\equiv-1\big(2\cdot4\cdot6...\cdot2k\big)^{p-1}\equiv-1\pmod{p}$$
So we have
$$\bigg(\frac{(\frac{p-1}{4})!}{p}\bigg)=\bigg(\frac{(2k)!}{p}\bigg)=-1$$
Which is exactly what we wanted to prove.
Thank you!
Wouldn't any Fermat prime $p = 2^{2^n} + 1$ for $n \ge 2$ be a counterexample?
We have $$\frac {p-1}{2^{v_2(p-1)-1}} = \frac {2^{2^n}}{2^{2^n-1}}=2$$
To find $\left(\dfrac{2!}p\right)$ we use the Second Supplement to Law of Quadratic Reciprocity:
$$p \equiv 1 \pmod 8 \implies \left(\frac2p\right) = 1$$
Indeed the smallest counterexample (should be) $17$, where $2 \equiv 6^2 \pmod {17}$.