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Let $a$, $b$, $c$ be natural numbers such that $$ a^2 +b^2=c^2 $$ and $$c-b=1.$$

Prove that $$a^b + b^a $$ is divisible by $c$. Any hints??

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  • $\begingroup$ $a^2=2b+1$$ WLOG $a=2d+1, b=?$ $\endgroup$ – lab bhattacharjee May 13 '18 at 14:31
  • $\begingroup$ @labbhattacharjee yes I did till that point , then what?? $\endgroup$ – Kavita Juneja May 13 '18 at 14:33
  • $\begingroup$ @labbhattacharjee yes I did till that point , then what?? B comes out as a multiple of 4.. $\endgroup$ – Kavita Juneja May 13 '18 at 14:42
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$a^2=2b+1=2(c-1)+1\equiv-1\pmod c$

$b=c-1\equiv-1,b^a\equiv-1$ as $a$ is odd

If $a=2d+1,b=4\cdot\dfrac{d(d+1)}2$

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