## Understand the problem

## Start with hints

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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.

##### Topic

##### Difficulty Level

8.5 out of 10

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I have a doubt in this question of 3 primes (ISI 2017).

While Proving that for m>9,there is no solution,the hint says that gap is >8.What does this mean and how does it contradict the fact that there can be solutions for m>9.

Yes, there was a mistake , I have corrected that. I think now it is clear to you. But more clearly I can say that the gap between two consecutive primes is a increasing function. This concept is used here.