Given that \(\quad a+b+c=0\)

Prove that : \[2(a^4+b^4+c^4)=(a^2+b^2+c^2)^2\]

Nice proofs are always welcome!

Given that \(\quad a+b+c=0\)

Prove that : \[2(a^4+b^4+c^4)=(a^2+b^2+c^2)^2\]

Nice proofs are always welcome!

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TopNewestLets start off by finding the value of \(\ a^2 +b^2 +c^2 \) in terms of another sum by doing the following: \[(a+b+c)^2 = a^2 +b^2 +c^2 +2(ab +ac +bc) = 0 \Rightarrow\ (a^2 +b^2 +c^2)^2 = 4(ab+ac+bc)^2 \]

Now expanding out the RHS: \[... = 4(a^2 b^2 +a^2 c^2 + b^2 c^2 +2abc(a+b+c)) = 4(a^2 b^2 +a^2 c^2 + b^2 c^2) \]

Now if we expand out \(\ (a^2 +b^2 +c^2)^2 \) and cancel out terms we get: \[ a^4 +b^4 +c^4 = 2 (a^2 b^2 +a^2 c^2 + b^2 c^2) \ ........ \ (1) \] \[\Rightarrow\ 2(a^4 +b^4 +c^4) = a^4 +b^4 +c^4 + 2 (a^2 b^2 +a^2 c^2 + b^2 c^2) \]

\[\therefore\ 2(a^4 +b^4 +c^4) = (a^2 +b^2 +c^2)^2 \] – Curtis Clement · 1 year, 9 months ago

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– Nihar Mahajan · 1 year, 9 months ago

Yeah! Thanks , I got an easier way . Read below.Log in to reply

Approach: Factoring polynomials using the (slightly abused) remainder factor theorem for polynomials in several degrees.

Define \( f(a, b, c) = 2 (a^4 + b^4 + c^4) - (a^2 + b^2 + c^2) ^2 \).

We want to show that if \( a+b+ c = 0 \), then \( f(a, b, c) = 0 \). This strongly suggests to us that \( f(a, b, c) = (a+b+c) \times g(a, b, c) \).

If \( a + b + c = 0 \), then we have \( f( -b-c, b, c ) = 0 \). Notice that the powers of \(a \) only occur in even degree, hence \( f(a, b, c) = f(-a, b, c) \). This implies that \( f( b+c, b, c ) = 0 \), and thus that \( a-b-c \mid f(a, b, c) \). Cyclically, we also get that \( b-c-a \mid f(a,b,c)\) and \( c-a-b \mid f(a,b,c) \).

Putting this all together, this strongly suggests that

\[ (a+b+c)(a-b-c)(b-c-a)(c-a-b) \mid f(a,b,c) \]

Since both sides have the same degree (4), we know that they differ by a constant. In fact, we can now verify that

\[ f(a, b, c) = - (a+b+c)(a-b-c)(b-c-a)(c-a-b) \] – Calvin Lin Staff · 1 year, 9 months ago

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The equation \(2(a^4 + b^4 + c^4) = (a^2 + b^2 + c^2)^2\) is equivalent to \[-a^4 - b^4 - c^4 + 2a^2 b^2 + 2a^2 c^2 + 2b^2 c^2 = 0.\] This equation factors as \[(a + b + c)(-a + b + c)(a - b + c)(a + b - c) = 0,\] and the result follows.

This factorization comes up in Heron's formula: \[ \begin{align*} K^2 &= s(s - a)(s - b)(s - c) \\ &= \frac{1}{16} (a + b + c)(-a + b + c)(a - b + c)(a + b - c) \\ &= \frac{1}{16} (-a^4 - b^4 - c^4 + 2a^2 b^2 + 2a^2 c^2 + 2b^2 c^2). \end{align*} \] – Jon Haussmann · 1 year, 9 months ago

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– Nihar Mahajan · 1 year, 9 months ago

Really great , I was curious of a heron's formula solution. Thanks!Log in to reply

– Calvin Lin Staff · 1 year, 9 months ago

Great! That's the "Use Heron's Formula" that I alluded to. There are many ways to write the expression of \(K\), and most people are only familiar with the factored form, instead of the expanded form.Log in to reply

Use the convention of \(S_1 = a+b+c\), \(S_2=ab+bc+ca\) and \(S_3 = abc\), and \(P_n = a^n+b^n+c^n\). Then by Newton's Sum method:

\(P_2 = S_1 P_1 - 2S_2 = 0 - 2S_2 = -2S_2\)

\(S_4 = S_1P_3-S_2P_2+S_3P_1 = 0 -S_2(-2S_2) + 0 = 2S_2^2\)

Now \(LHS = 2P_4 = 4S_2^2\) and \(RHS = P_2^2 = (-2S_2)^2 = 4S_2^2 = LHS\) – Chew-Seong Cheong · 1 year, 9 months ago

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Do you know how to get started? – Calvin Lin Staff · 1 year, 9 months ago

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– Nihar Mahajan · 1 year, 9 months ago

I am using Newton's identities , but am unable to get the desired result. :(Log in to reply

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help here too. Thanks in advance. – Nihar Mahajan · 1 year, 9 months ago

Sir ,Log in to reply

\[\sigma_1=a+b+c=0 \\ \sigma_2=ab+bc+ac \\\sigma_3=abc\\ S_1=a+b+c=0 \\ S_2=a^2+b^2+c^2 \\S_3=a^3+b^3+c^3\]

By Newton's identity, \[S_2=\sigma_1^2-2\sigma_2 \\ \sigma_2= -\dfrac{S_2}{2}\]

By Newton's identity , \[S_4=\sigma_1S_3 - \sigma_2S_2+\sigma_3S_1 \\ S_4 = 0 -\left(\dfrac{-S_2}{2}\right)S_2+0 \\ 2S_4=S_2^2\]

Hence , proved.

I had forgotten that \(S_1 = \sigma_1\) which caused problem.Sorry Calvin Sir for wasting your valuable time for such straightforward problem.Also , your approach is also good and I will try it too.

Should I delete this note , since my problem is solved?Thanks! – Nihar Mahajan · 1 year, 9 months ago

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– Sanjeet Raria · 1 year, 9 months ago

Wow! So you're going to use this result in one of your next questions?Log in to reply

– Nihar Mahajan · 1 year, 9 months ago

Nope , this was a question in one of my exercises! Due to many variables I had forgotten that \(S_1=\sigma_1\) which led me out of correct way.Log in to reply

– Sanjeet Raria · 1 year, 9 months ago

Okay. .. keep it up! :)Log in to reply

Leave this note up, so that others can refer to it.

Think about the other approaches that I listed, and see how they work. – Calvin Lin Staff · 1 year, 9 months ago

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– Nihar Mahajan · 1 year, 9 months ago

Indeed , this is a beautiful result. I will surely look forward for more approaches.Thanks a lot for your co-operation.Log in to reply

Plug In 1+1-2 xD xD Just kidding. I'm on it! – Mehul Arora · 1 year, 9 months ago

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– Nihar Mahajan · 1 year, 9 months ago

That's the most special case. -_-Log in to reply