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: \[ a^4 +b^4 +c^4 = 2 (a^2 b^2 +a^2 c^2 + b^2 c^2) \ ........ \ (1) \] \[\Rightarrow\ 2(a^4 +b^4 +c^4) = a^4 +b^4 +c^4 + 2 (a^2 b^2 +a^2 c^2 + b^2 c^2) \]

\[\therefore\ 2(a^4 +b^4 +c^4) = (a^2 +b^2 +c^2)^2 \]

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Yeah! Thanks , I got an easier way . Read below.

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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

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

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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: \[ \begin{align*} K^2 &= s(s - a)(s - b)(s - c) \\ &= \frac{1}{16} (a + b + c)(-a + b + c)(a - b + c)(a + b - c) \\ &= \frac{1}{16} (-a^4 - b^4 - c^4 + 2a^2 b^2 + 2a^2 c^2 + 2b^2 c^2). \end{align*} \]

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

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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.

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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 \(RHS = P_2^2 = (-2S_2)^2 = 4S_2^2 = LHS\)

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Comment deleted Apr 18, 2015

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Seems pretty straightforward to me. What are you struggling with?

Do you know how to get started?

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

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help here too. Thanks in advance.

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

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!

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

Think about the other approaches that I listed, and see how they work.

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Plug In 1+1-2 xD xD Just kidding. I'm on it!

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That's the most special case. -_-

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