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Editorial
$(r^2-q)(r^2+q)=p $-->$ r^2-q=1,r^2+q=p$ --> $2q=p-1 , 2r^2=p+1 $
if $3|r \rightarrow 3 = r \rightarrow p = 17 \rightarrow q = 8$ which is not prime.
Determine the sum of all prime numbers $p, q$ and $r$ with
$p + q^2 = r^4$
$(r^2-q)(r^2+q)=p $-->$ r^2-q=1,r^2+q=p$ --> $2q=p-1 , 2r^2=p+1 $
if $3|r \rightarrow 3 = r \rightarrow p = 17 \rightarrow q = 8$ which is not prime.