 Let $a, b, c$ are natural numbers 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
Notice that $a^2 = c^2 – b^2 = (c+b)(c-b)$

But c – b = 1. Hence $a^2 = c + b$. But c and b are consecutive numbers (after all their difference is 1!). Sum of two consecutive numbers is always odd (Why?).

Hence $a^2$ is odd. This implies a is odd.

Replace c by 1+b. We have $a^2 + b^2 = ( 1+ b)^2 = 1 + b^2 + 2b$

But that means $a^2 = 1 + 2b$. This implies $2b = a^2 – 1 = (a-1)(a+1)$

We already showed a is odd. Hence odd + 1, odd -1 are consecutive even numbers. Hence atleast one of them must be divisible by 4 implying their product must be divisible by 8 (as the other is divisible by at least 2).

Hence 2b equals something that is divisible by 8. This implies b is divisible by 4.

We already found that $a^2 = c + b$. We test this modulo c (Modular Arithmetic is a useful tool from Number Theory that you should definitely learn).

$a^2 \equiv c + b \equiv b \mod c$ . But since b = c – 1, hence $b \equiv -1 \mod c$. Hence $a^2 \equiv -1 \mod c$

Now recall that b is divisible by 8. Hence b/2 is even. This implies $(a^2)^{\frac{b}{2}} \equiv (-1)^{\frac{b}{2}} = 1 \mod c$. Hence $a^b \equiv 1 \mod c$

On the other hand a is odd and b = c-1. Hence $b^a \equiv (-1)^a \equiv -1 \mod c$

Adding we get the final answer $a^b + b^a \equiv 1 – 1 \equiv 0 \mod c$

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