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Division is somewhat more complex than I thought

Let \(a, b, c\) be all positive integers such that \(a^2 +b^2= c^2 \).

Prove that, for all positive integers \(n\), both \(a^{2n+1}+b^{2n+1}+c^{2n+1}\) and \((b+c)(c+a)(a+b)\) are divisible by \(a+b+c\).
Hey guy S please give your comments and also your efforts to solve the problem as I am waiting eagerly of any comments of brillianters

Note by Biswajit Barik
7 months, 2 weeks ago

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First let's show that \(ab\) is divisible by \(a+b+c\):

\[ab = \dfrac{(a+b)^2-a^2-b^2}{2} = \dfrac{(a+b)^2 - c^2}{2} = \dfrac{(a+b+c)(a+b-c)}{2}\]

Now \(a+b+c\) and \(a+b-c\) differ by an even number, so they have the same parity. However, they can't both be odd, because then the final expression above would not be an integer, even though it's equal to \(ab\) which is an integer. Therefore \(a+b-c\) is even, so \(\dfrac{a+b-c}{2}\) is an integer. Therefore \(ab\) is divisible by \(a+b+c\). \(\square\)

For the rest of the proof, for brevity let \(p = a+b+c\) and let \(q = \dfrac{ab}{a+b+c}\) (which is an integer by the above lemma).

\[\begin{align} (b+c)(c+a)(a+b) & = (p-a)(p-b)(p-c) \\ & = p^3 - (a+b+c)p^2 + (ab+ac+bc)p - abc \\ & = (ab+ac+bc)p - pqc \\ (a+b)(b+c)(c+a) & = p(ab+ac+bc - qc) \end{align}\]

Therefore \((b+c)(c+a)(a+b)\) is divisible by \(p\). \(\square\)

Next we'll show that \(a^{2n+1} + b^{2n+1} + c^{2n+1}\) is divisible by \(p\) for any \(n \ge 0\):

\[a^{2n+1} + b^{2n+1} + c^{2n+1} = a^{2n+1} + b^{2n+1} - (a+b)^{2n+1} + \left[(a+b)^{2n+1} + c^{2n+1} \right] \]

I say that \((a+b)^{2n+1} + c^{2n+1}\) is divisible by \(p\). Why? Using modular arithmetic, we have \(a+b \equiv -c \pmod{p}\). Therefore, \[(a+b)^{2n+1} + c^{2n+1} \equiv (-c)^{2n+1} + c^{2n+1} \equiv -c^{2n+1} + c^{2n+1} \equiv 0 \pmod{p}\]

Hence \((a+b)^{2n+1} + c^{2n+1} = pr\) for some integer \(r\). Continuing with the proof:

\[\begin{align} a^{2n+1} + b^{2n+1} + c^{2n+1} & = a^{2n+1} + b^{2n+1} - (a+b)^{2n+1} + \left[(a+b)^{2n+1} + c^{2n+1} \right] \\ & = a^{2n+1} + b^{2n+1} - (a+b)^{2n+1} + pr \\ & = pr + a^{2n+1} + b^{2n+1} - \sum_{k=0}^{2n+1} \binom{2n+1}{k} a^{2n+1-k} b^k \\ & = pr - \sum_{k=1}^{2n} \binom{2n+1}{k} a^{2n+1-k} b^k \end{align}\]

Now let's consider that last sum. For each \(k\) between \(1\) and \(2n\), the exponents of \(a\) and \(b\) are always at least \(1\). Therefore for each \(k\), \(a^{2n+1-k} b^k\) is divisible by \(ab\). Hence the entire sum is divisible by \(ab\). Let's say the sum equals \(abs\) for some integer \(s\).

\[\begin{align} a^{2n+1} + b^{2n+1} + c^{2n+1} & = pr - abs \\ & = pr - pqs \\ a^{2n+1} + b^{2n+1} + c^{2n+1} & = p(r-qs) \end{align}\]

Therefore, \(a^{2n+1} + b^{2n+1} + c^{2n+1}\) is divisible by \(p\). \(\square\) Ariel Gershon · 4 months, 1 week ago

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Hello sir try another Problem coordinates are Also tough Biswajit Barik · 4 months, 1 week ago

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Sir could you make me understand what poincare conjecture is i tried IT but i van Not Figure out Biswajit Barik · 4 months, 1 week ago

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Although your answer is correct Biswajit Barik · 4 months, 1 week ago

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Thank you ariel Sir but why do you Follow me could you tell me for what reason Biswajit Barik · 4 months, 1 week ago

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Superb Ariel Biswajit Barik · 4 months, 1 week ago

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What have you tried? Where are you stuck? Calvin Lin Staff · 7 months, 1 week ago

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