Let$ p_1, p_2, p_3$ be primes with $p_2 \neq p_3$, such that $4 + p_1p_2$ and $4 + p_1p_3$ are perfect squares. Find all possible values of$ p_1, p_2, p_3$.
My approach:
Assume $p_2>p_3$ and let $4+p_1p_2=m^2$ and $4+p_1p_3=k_2$ $$ \implies m^2−4=(m−2) (m+2)=p_1p_2$$ and, $$ k^2−4=(k−2)(k+2)=p_1p_3$$
-> $p_1=m−2$ or $m+2$ and $p_1=k−2$ or $k+2$.
If $p_1=m−2$ then $p_2=m+2$. Then if $p_1=k−2$ then $k=m$ and $p_2=p_3$ which is impossible. And if $p_1=k+2$ then $p_3=k−2<p_1<p_2$ a contradiction.
Hence, $p_1=m+2$ and $p_2=m−2$. If $p_1=k+2$ we get $m=k$ and $p_2=p_3$ so $p_1=k−2$ and $p_3=k+2$.
$$ \implies p_3=p_1+4=p_2+8$$
If $p_2=2$ then $p_1,p_3$ are even which is impossible.
I am stuck now.
Hint: Show that $p_3=p_1+4=p_2+8$ implies that one of $p_1,p_2,p_3$ is divisible by 3.