Let 4+p1p2=m2. So m2−4=(m−2)(m+2)=p1p2
Let 4+p1p3=k2. So k2−4=(k−2)(k+2)=p1p3.

So p1=m−2 or m+2 and p1=k−2 or k+2
If p1=m−2 then p2=m+2. Then if p1=k−2 then k=m and p2=p3 which is impossible. And if p1=k+2 then p3=k−2<p1<p2 a contradiction.

So p1=m+2 and p2=m−2. If p1=k+2 we get m=k and p2=p3 so p1=k−2 and p3=k+2.

So p3=p1+4=p2+8.

If p2=2 then p1,p3 are even which is impossible.

If p2=3 then p1=7;p3=11 and 4+p1p2=52;4+p1p3=92.

If p2>3 then p2=p1−4≡p1−1mod3.

p3=p1+4≡p1+1mod3.

If p1≡0mod3 then 3|p1 which is a contradiction.

If p1≡1mod3 then p2≡0mod3 which is a contradiction as p_2 is prime.

Likewise if p1≡−1mod3 then p3≡0mod3 which is also impossible.

So p2>3 is impossible and so the only solutions are {p2,p3}={3,11};p1=7.

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## Comments

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TopNewestWolog assume p2<p3.

Let 4+p1p2=m2. So m2−4=(m−2)(m+2)=p1p2 Let 4+p1p3=k2. So k2−4=(k−2)(k+2)=p1p3.

So p1=m−2 or m+2 and p1=k−2 or k+2 If p1=m−2 then p2=m+2. Then if p1=k−2 then k=m and p2=p3 which is impossible. And if p1=k+2 then p3=k−2<p1<p2 a contradiction.

So p1=m+2 and p2=m−2. If p1=k+2 we get m=k and p2=p3 so p1=k−2 and p3=k+2.

So p3=p1+4=p2+8.

If p2=2 then p1,p3 are even which is impossible.

If p2=3 then p1=7;p3=11 and 4+p1p2=52;4+p1p3=92.

If p2>3 then p2=p1−4≡p1−1mod3.

p3=p1+4≡p1+1mod3.

If p1≡0mod3 then 3|p1 which is a contradiction.

If p1≡1mod3 then p2≡0mod3 which is a contradiction as p_2 is prime.

Likewise if p1≡−1mod3 then p3≡0mod3 which is also impossible.

So p2>3 is impossible and so the only solutions are {p2,p3}={3,11};p1=7.

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Thanks :)

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could you please provide detailed solution to this

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3,7,11

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