Prime number!

Let p1,p2,p3 be primes with p2≠p3 such that 4+p1p2 and 4+p1p3 are perfect squares. Find all possible values of p1,p2,p3.

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`2 \times 3`	$2 \times 3$
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Comments

Wolog 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.

Tarit Goswami - 3 years, 3 months ago

Thanks :)

Santosh Jaju - 3 years, 3 months ago

3,7,11

Krishan Yadav - 3 years, 4 months ago

could you please provide detailed solution to this

Santosh Jaju - 3 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$