6. Given a prime number $p$ satisfying $p \equiv-1(\bmod 4), n$ is a positive integer. Then “ $2 \sqrt{1+4 p n^{2}}$ is a positive integer” is a “ $2+$ $2 \sqrt{1+4 p n^{2}}$ is a perfect square” $\qquad$ condition (fill in “sufficient but not necessary”, “necessary and sufficient”, or “necessary but not sufficient”).


6. Necessary and Sufficient.

Necessity is evident.
Sufficiency: From $2 \sqrt{1+4 p n^{2}}=m(m \in \mathbf{N})$, we have $1+4 p n^{2}=\frac{m^{2}}{4} \in \mathbf{N}$.
Also, $1+4 p n^{2}=(2 t+1)^{2}(t \in \mathbf{N})$, so $p n^{2}=t(t+1)$.
Since $(t, t+1)=1$, one of $t$ or $t+1$ must be a perfect square, and the other must be a perfect square multiplied by $p$.
If $t+1=p u^{2}, t=v^{2}(u, v \in \mathbf{N})$, then $p u^{2}=v^{2}+1 \equiv 1$ or $2(\bmod 4)$,

but $p u^{2} \equiv(-1) \times u^{2} \equiv 0$ or $3(\bmod 4)$, which is a contradiction. Therefore, $t+1=u^{2}, t=p v^{2}(u, v \in \mathbf{N})$. Hence, $2+2 \sqrt{1+4 p n^{2}}=4(t+1)=4 u^{2}=(2 u)^{2}$.