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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