Given that $a$, $b$, $c$ are natural numbers, with $a^2+b^2=c^2$ and $ c-b=1$, prove the following
- $a$ is odd
- $b$ is divisible by 4
- $a^b + b^a$ is divisible by $c$
My approach to prove the first statement is as follows: given that $a² + b² = c²$:
$$a^2 = (c^2 – b^2)$$
$$a^2 = (c + b)(c – b)$$
Given that $(c – b) = 1$,
$$a^2 = c + b = 2c – 1$$
This implies $a^2$ is odd, which implies(from some established trivial result I remember) that a is odd.
For the second part, I figured out that either b or c must be odd, given that they are consecutive natural numbers. Having proved $a^2$ is odd, I suspect $c^2$ must be odd(absolutely out of intuition and vague reasoning that I'll mention in the end). I have no idea how to proceed beyond that.
I'm absolutely clueless about the third part, and I feel it concerns Number Theory, something I'm not familiar with yet.
My intuition: a, b, c are Pythagorean triplets such as $(3, 4, 5)$, $(5, 12, 13)$ and $(7, 24, 25)$; I feel many more exist. I would like an explanation behind these patterns too.
My background: I'm in the last year of high school; I can comprehend basic theoretical proofs, and have little idea about number theory. The above question is from an undergrad entrance exam, intended for high school passouts.
I sincerely apologise for not using MathJax yet again; every time I try to use it I end up getting confused. I assure you I'll learn it in the time to come 🙂
Best Answer
Your first part is correct. We have that $b=c-1$ and therefore $$a^2=c^2-b^2=c^2-(c-1)^2=2c-1$$ which implies that $a$ is odd.
Now the third part follows as soon as you solve the second part: if $b$ is divisible by $4$ then $b=4k$ and $$a^b+b^a=(2c-1)^{2k}+(c-1)^{2c-1}\equiv (-1)^{2k}+(-1)^{2c-1}=1-1=0\pmod{c}.$$