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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]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 \(p_2\neq p_3\) \(\Rightarrow m\neq n\).
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]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,\( p_1=7,p_2=3,p_3=11\) .
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.7"]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 \(p_1=7,p_2=11,p_3=3\).
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]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.
[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2017. Subjective Problem no. 6.
8.5 out of 10
[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.3.4" hover_enabled="0" open="off"]Elementary Number Theory by David M. Burton [/et_pb_accordion_item][/et_pb_accordion][et_pb_blurb title="I.S.I. & C.M.I. Entrance Program" image="https://cheenta.com/wp-content/uploads/2018/03/ISI.png" _builder_version="3.22.4" header_level="h1" header_font="||||||||" header_text_color="#e02b20" header_font_size="50px" body_font="||||||||"]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.
The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.
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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.