Pythagorean Triple, ISI Subjective Entrance 2018, Problem 7

Understand the problem

'Elementary Number Theory' by David M. Burton
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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).

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

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

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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