ISI Power

If $p_3\ne p_2$

Let $\begin{cases}p_1p_2 + 4 =a^2 \\ p_1p_3 + 4 =b^2 \end{cases}$

Thus

$p_1p_2 = a^2 - 4 = a^2 - 2^2 = (a+2)(a-2)$

$p_1p_3 = b^2 - 4 = b^2 - 2^2 = (b+2)(b-2)$

$case(i)$ $p_1=a+2$ & $p_2=a-2$

$case(ii)$ $p_1=a-2$ & $p_2=a+2$

$case(iii)$ $p_1=b+2$ & $p_3=b-2$

$case(iv)$ $p_1=b-2$ & $p_3=b+2$

Now $case(i)$ & $case(iii)$ can't be true simultaneously as that would imply $p_2 = p_3$ .

Similarly $case(ii)$ & $case(iv)$ can't be true simultaneously either.

Thus let us consider without any loss of generality $b>a \Rightarrow p_3 > p_2$ .

That would imply $case(i)$ & $case(iv)$ are true.

Thus $p_1 = a+2 = b-2$

$p_2=a-2 = p_1-4$

$p_3=b+2 = p_1+4$

Now we have to find $p_1$ such that $p_1-4,p_1,p_1+4$ all are primes.

Now all primes are of the form $6k+1$ or $6k-1$

Thus if $p_1 = 6k-1$ then

$p_2 = 6k-1-4 =6k-5=6(k-1)+1$ thus it can be a prime.

But $p_3 = 6k-1+4=6k+3$ thus it is a multiple of $3$ and thus not a prime.

Then if $p_1=6k+1$ then

$p_3 = 6k+1+4 =6k+5=6(k+1)-1$ thus it can be a prime.

and $p_2=6k+1-4=6k-3$ thus it is a multiple of three and not a prime, except when $k=1$ then $p_2=3$ which is a prime.

Therefore, If $k=1$ then

$p_1=6(1)+1=7$

$p_2=6(1)-3=3$

$p_3=6(1+1)-1=11$

And if $p_2>p_3$ then

$p_1=6(1)+1=7$

$p_2=6(1+1)-1=11$

$p_3=6(1)-3=3$

Thus the possible solutions for $(p_1,p_2,p_3)$ are

$(7,3,11)$ & $(7,11,3)$

Thus number of solutions is $\boxed{2}$

Note : As I mentioned in the beginning, this proof is based on the fact that $p_2 \ne p_3$

If $p_2=p_3$ was possible then any pair of cousin primes (i.e. primes with difference of 4) will be a valid solution. And we don't know how many of those exist (yet).

Proof that all primes are of the form $6k \pm 1$ .

All numbers can be expressed as one of these,

$6k,6k+1,6k+2,6k+3,6k+4,6k+5$

Out of which $6k$ is divisible by $6$ .

$6k+2$ is divisible by $2$ .

$6k+3$ is divisible by $3$ .

Thus if a number is prime then it must be of form $6k+1$ or $6k+5=6(k+1)-1=6k'-1$

However the converse is not true,

Example:

$43$ is a prime thus $43 = 6(7)+1$

However, $35 = 6(6)-1 = 5 \times 7$ thus $35$ is not a prime.