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Pythagorean Triple, ISI Subjective Entrance 2018, Problem 7

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Understand the problem

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\(a^2+b^2=c^2\) and \(c-b=1\).
Prove that
(i) \(a\) is odd,
(ii) \(b\) is divisible by 4,
(iii) \(a^b+b^a\) is divisible by \(c\).  

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I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2018. Subjective Problem no. 7.
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Number Theory

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]8 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
'Challenge and Thrill of Pre-College Mathematics' by V,Krishnamurthy, C.R.Pranesachar, ect. [/et_pb_accordion_item][/et_pb_accordion][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" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

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Do you really need a hint? Try it first!

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\(a^2+b^2=c^2=(b+1)^2=b^2+2b+1\) \(\Rightarrow a^2=2b+1\) Which implies \(a^2\) is odd integer. \(\Rightarrow a\) is also an odd integer =\(2k+1\) (say).

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\((2k+1)^2+b^2=c^2\) \(\Rightarrow 4k^2+4k+1+b^2=b^2+2b+1\) \(\Rightarrow b=2k(k+1)\). Now \(k(k+1) \) is always even =\(2l\) (say). Therefore, \(b=4l\), i.e. \(b\) is divisible by 4.  

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\(a^b+b^a=a^{4l}+b^{2k+1}=(a^2)^{2l}+(c-1)^{2k+1}\) =\((2b+1)^{2l}+(c-1)^{2k+1}\) =\((2c-1)^{2l}+(c-1)^{2k+1}.\)

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Now \(2l\) is even , therefore , \((2c-1)^{2l} \) is of the form : \( 2cp+1\) where \(p \in \mathbb{N}\). And \(2k+1\) is odd , therefore \((c-1)^{2k+1} \) is of the form : \(cq-1\), where \(q \in \mathbb{N}\). Therefore \(a^b+b^a=(2cp+1)+(cq-1)=c\cdot (2p+q)\). \(\Rightarrow c|a^b+b^a\).

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Connected Program at Cheenta

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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.

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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Similar Problem

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