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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 :)

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5 Answers 5

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

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  • $\begingroup$ Thankyou, I believe the third part assumes knowledge of modular arithmetic? $\endgroup$
    – Manan
    Jan 26, 2020 at 9:43
  • $\begingroup$ 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 $c$ $\endgroup$
    – Robert Z
    Jan 26, 2020 at 9:47
  • $\begingroup$ Thanks again. I'll look into elementary number theory and modular arithmetic. $\endgroup$
    – Manan
    Jan 26, 2020 at 9:49
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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.

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  • $\begingroup$ Shouldn't c = b + 1? $\endgroup$
    – Manan
    Jan 26, 2020 at 10:01
  • $\begingroup$ @MananJain Yes, sorry, fixed it. $\endgroup$
    – Miguel
    Jan 26, 2020 at 10:05
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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 $4$.

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  • $\begingroup$ Thankyou very much :) $\endgroup$
    – Manan
    Jan 26, 2020 at 9:42
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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.

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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.

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

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