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 \(p_1,p_2,p_3\).   

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.   

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Source of the problem

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

Topic
Number Theory

Difficulty Level

8.5 out of 10

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