 ## Understand the problem

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 $p_1,p_2,p_3$.

Do you really need a hint? Try it first!

Let $4+p_1p_2=m^2$ and $4+p_1p_3=n^2$, where $m,n \in \mathbb{N}$. $\Rightarrow p_1p_2=(m-2)(m+2)$ and $p_1p_3=(n+2)(n-2)$. Since $p_1,p_2,p_3$ are primes with $p_2\neq p_3$ $\Rightarrow m\neq n$.

Case –I: $p_2<p_3 \Rightarrow m<n$. Clearly, $p_1=m+2=n-2, \Rightarrow n=m+4$              $p_2=m-2$    and                $p_3=n+2=m+6$. Therefore, $(m+2),(m-2),(m+6)$ are all prime numbers.               We see that $m=5$ satisfy the above condition. And then,$p_1=7,p_2=3,p_3=11$ .

Case-II: $p_2>p_3 \Rightarrow m>n$. $\Rightarrow p_1=m-2=n+2 \Rightarrow m=n+4$                        $p_2=m+2=n+6$   and                          $p_3=n-2$. Now $(n+2),(n+6), (n-2)$ all are primes. Again , $n=5$ satisfy this condition. Hence $p_1=7,p_2=11,p_3=3$.

Thus all possible values of $p_1,p_2,p_3$ are ( 7,3,11)  and (7,11,3). Now need to conclude that there does not exist any more triple of prime numbers satisfying the given condition. Consider these numbers:            $p_1=m+2,p_2=m-2,p_3=m+6$ , now the gaps between $p_1,p_2,p_3$ are given by:                 $p_1-p_2=4, p_3-p_2=8$  and  $p_3-p_1=4$. We see that for $m>9$ these three gaps cannot be 4,8 and 4 simultaneously . That is at least one of these three gaps is greater than 4 for $m>9$ . And between 1 to 9 only $m=5$ satisfy the given condition. Hence there does not exist any more triples.

## Connected Program at Cheenta

##### Source of the problem

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2017. Subjective Problem no. 6.

Number Theory

8.5 out of 10

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