# Understand the problem

Let \(a,b,c \in \mathbb{N}\) be such that

\(a^2+b^2=c^2\) and \(c-b=1\).

Prove that

(i) \(a\) is odd,

(ii) \(b\) is divisible by 4,

(iii) \(a^b+b^a\) is divisible by \(c\).

##### Source of the problem

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2018. Subjective Problem no. 7.

##### Topic

Number Theory

##### Difficulty Level

8 out of 10

##### Suggested Book

‘Elementary Number Theory’ by David M. Burton

‘Challenge and Thrill of Pre-College Mathematics’ by V,Krishnamurthy, C.R.Pranesachar, ect.

# Start with hints

Do you really need a hint? Try it first!

\(a^2+b^2=c^2=(b+1)^2=b^2+2b+1\)

\(\Rightarrow a^2=2b+1\) 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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