Solution:
$p\neq 5$-->$(5,p)=1$
$5^{\phi (p^{2})} \equiv 1 \pmod{p^{2}}$
$5^{p^{2}-p} \equiv 1 \pmod{p^{2}}$
$5^{p^{2}}+1 \equiv 5^{p}+1 \equiv 0 \pmod{p^{2}}$
$p^{2}k =5^{p}+1,k \in \mathbb{N}---(*)$
$\because p \in \mathbb{P}$
$5^{p-1} \equiv 1 \pmod{p}$
$5^{p} \equiv 5 \pmod{p}$--->$p \neq 2$
$5^{p}=pm+5,m \in \mathbb{N}---(**)$
from$(*),(**)$
$6=p(pk-m)$
given $pk-m = r,r \in \mathbb{N}$
$pr=6$
$\therefore p =3$
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