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

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.

Topic

Number Theory

Difficulty Level

8.5 out of 10

Suggested Book

Elementary Number Theory by David M. Burton

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

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