Factorial and square number

$p!+1= (2p+1)^{2}$

$p!+1=4p^{2}+4p+1$

$p!=p(4p+4)$

$(p-1)!=4p+4$

$(p-1)! \equiv 4p+4 \equiv -1 \pmod {p}$ (By Wilson's theorem)

$4p \equiv -5 \pmod {p}$

$0 \equiv -5 \pmod {p}$

$5 \equiv 0 \pmod {p}$

$\Rightarrow p=5$

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$q!+1=(10q+5-4)^{2}$

$q!+1=(10q+1)^{2}$

$q!+1=100q^{2}+20q+1$

$q! = q(100q+20)$

$(q-1)! = 100q+20$

$(q-1)! \equiv 100q+20 \equiv -1 \pmod {q}$ (By Wilson's theorem)

$100q \equiv -21 \pmod {q}$

$0 \equiv -21 \pmod {q}$

$21 \equiv 0 \pmod {q}$

$\Rightarrow q=3 (rej.)$ or $q=7$

$\Rightarrow q=7$

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$p+q = 5+7 =12$