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## 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.
Number Theory

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.

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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