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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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