## Understand the problem

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

##### Topic

##### Difficulty Level

##### 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 \(a^b+b^a=(2cp+1)+(cq-1)=c\cdot (2p+q)\). \(\Rightarrow c|a^b+b^a\).

## Connected Program at Cheenta

# I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelorâ€™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. CheentaÂ offers an intense, problem-driven program for these two entrances.

Google