# Given that $a$, $b$, $c$ are natural numbers, with $a^2+b^2=c^2$ and $c-b=1$, prove the following.

Given that $$a$$, $$b$$, $$c$$ are natural numbers, with $$a^2+b^2=c^2$$ and $$c-b=1$$, prove the following

1. $$a$$ is odd
2. $$b$$ is divisible by 4
3. $$a^b + b^a$$ 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 :)

## 5 Answers

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 $$a^b+b^a=(2c-1)^{2k}+(c-1)^{2c-1}\equiv (-1)^{2k}+(-1)^{2c-1}=1-1=0\pmod{c}.$$

• Thankyou, I believe the third part assumes knowledge of modular arithmetic? Jan 26, 2020 at 9:43
• Yes, but it is very elementary. You just need that the remainder of the division of $(Ac-1)^m$ by $c$ is $(-1)^m$ i.e. $(Ac-1)^m-(-1)^m$ is divisible by $c$. Just expand $(Ac-1)^m$ and you will see that all the terms but the last one are divisible $c$ Jan 26, 2020 at 9:47
• Thanks again. I'll look into elementary number theory and modular arithmetic. Jan 26, 2020 at 9:49

Substituting $$c=b+1$$ we get $$a^2+b^2=b^2+2b+1 \iff 2b=a^2-1 \iff 2b=(a+1)(a-1)$$ Now if $$(a-1)=2k$$ and $$(a+1)=2l$$ where k and l are odd, then $$(a+1)-(a-1)=2$$ so $$2(l-k)=2$$ and $$l-k=1$$ which is a contradiction (they are both odd), therefore at least one of $$(a-1)$$ or $$(a+1)$$ is divisible by $$4$$ and $$b$$ is divisible by $$4$$.

For the third part, just note that $$a^2=-(c-1)^2+c^2=2c-1$$ which leaves remainder $$c-1$$ when divided by $$c$$, therefore $$a^k$$ leaves remainder $$1$$ when divided by $$c$$ for every positive even $$k$$ and since $$b=c-1$$ leaves remainder $$c-1$$, $$b^k$$ leaves remainder $$c-1$$ for every positive odd $$k$$ (you can see this by looking at the coefficient of $$(c-1)^k$$ that doesn't have a factor of $$c$$ in it). Therefore $$a^k+b^l$$ is divisible by $$c$$ for any positive even $$k$$ and odd positive $$l$$ since you proved $$a$$ is odd and $$b$$ is even the result follows.

• Shouldn't c = b + 1? Jan 26, 2020 at 10:01
• @MananJain Yes, sorry, fixed it. Jan 26, 2020 at 10:05

Your first point is excellent.

For point number 2, you do indeed get that $$c$$ must be odd (the sum of two odd squares is an even number that's not divisible by $$4$$, so it cannot be a square, thus $$a$$ and $$b$$ cannot both be odd). So we get $$b^2=(c-a)(c+a)$$ which, since $$a$$ and $$c$$ are both odd, is the product of two even numbers. Moreover, the difference between these two even numbers is $$2a$$, which is not divisible by $$4$$. Thus one of the two even numbers $$c-a$$ and $$c+a$$ is divisible by $$4$$. So $$b^2$$ is divisible by $$8$$ and, being a square, must therefore be divisible by $$16$$, which makes $$b$$ divisible by $$4$$.

• Thankyou very much :) Jan 26, 2020 at 9:42

The thing is b=(c-1) and a²=(2c-1) which is implies 2c-1 is a perfect square. 2c-1 = (2k+1)²(as a is odd).

Implies c = 2k(k+1) + 1 , as k(k+1) is divisible by 2 , implies c is of the form 4w+1. Which implies b=4w.

$$b^a\ +\ a^b$$ = $$(4w)^{\sqrt{8w+1}}\ +\ (8w+1)^{2w}$$

Now $$([4w+1]-1)^{odd\ number}$$ gives remainder -1 when divided 4w+1(by binomial theorem).

Same way $$(2[4w+1] - 1)^{even\ number}$$ gives remainder 1 when divided by 4w+1 (binomial theorm).

Hence $$a^b\ +\ b^a$$ gives remainder 0 when divided by 4w+1 hence divisible by 4w+1 which is nothing but c.

I didn't remember the exact approach, I did in the exam , I mean the problem appeared in ISI entrance 2018. However here is the solution I did in the exam. For part a)$$c-b=1$$ thus $$a^{2}=b+c$$ as $$b,c$$ are integers with opposite parity $$a^{2}$$ must be odd implying $$a$$ odd. For part b)Note that is $$c-b\equiv a^{2} \equiv b+c\equiv 1\mod 8$$ adding by $$-(c-b)$$ thus $$2b\equiv 0 \mod 8$$ Thus $$4$$ divides $$b$$ For part c)From now on, consider every equality $$\mod c$$ from now as From part b we have $$b=4k$$ thus $$(a^{2})^{2k}+b^{a}=(2c-1)^{2k}+(c-1)^{a}=0$$ as $$2k$$ and $$a$$ are of opposite parity and we are done.

• True, the question actually reappeared in the 2019 CMI entrance exam. Thanks for the answer as well :) Jan 28, 2020 at 18:29
• I don't think so CMI question paper was uploaded on the website and I didn't find this question.
– user410845
Jan 28, 2020 at 18:33
• cmi.ac.in/admissions/syllabus.php Jan 28, 2020 at 18:35
• Check out the 2019 question paper on the above link. Jan 28, 2020 at 18:35