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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