A nice result.

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!

Note by Nihar Mahajan
3 years, 9 months ago

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Lets 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: $a^4 +b^4 +c^4 = 2 (a^2 b^2 +a^2 c^2 + b^2 c^2) \ ........ \ (1)$ $\Rightarrow\ 2(a^4 +b^4 +c^4) = a^4 +b^4 +c^4 + 2 (a^2 b^2 +a^2 c^2 + b^2 c^2)$

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

- 3 years, 9 months ago

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

- 3 years, 9 months ago

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

$f(a, b, c) = - (a+b+c)(a-b-c)(b-c-a)(c-a-b)$

Staff - 3 years, 9 months ago

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*}

- 3 years, 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.

Staff - 3 years, 9 months ago

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

- 3 years, 9 months ago

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

- 3 years, 9 months ago

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

- 3 years, 9 months ago

That's the most special case. -_-

- 3 years, 9 months ago