A nice result.

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

Prove that : 2(a4+b4+c4)=(a2+b2+c2)22(a^4+b^4+c^4)=(a^2+b^2+c^2)^2

Nice proofs are always welcome!

Note by Nihar Mahajan
6 years ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

Lets start off by finding the value of  a2+b2+c2\ a^2 +b^2 +c^2 in terms of another sum by doing the following: (a+b+c)2=a2+b2+c2+2(ab+ac+bc)=0 (a2+b2+c2)2=4(ab+ac+bc)2(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(a2b2+a2c2+b2c2+2abc(a+b+c))=4(a2b2+a2c2+b2c2)... = 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  (a2+b2+c2)2\ (a^2 +b^2 +c^2)^2 and cancel out terms we get: a4+b4+c4=2(a2b2+a2c2+b2c2) ........ (1) a^4 +b^4 +c^4 = 2 (a^2 b^2 +a^2 c^2 + b^2 c^2) \ ........ \ (1)  2(a4+b4+c4)=a4+b4+c4+2(a2b2+a2c2+b2c2)\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)

 2(a4+b4+c4)=(a2+b2+c2)2\therefore\ 2(a^4 +b^4 +c^4) = (a^2 +b^2 +c^2)^2

Curtis Clement - 6 years ago

Log in to reply

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

Nihar Mahajan - 6 years ago

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(a4+b4+c4)(a2+b2+c2)2 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 a+b+ c = 0 , then f(a,b,c)=0 f(a, b, c) = 0 . This strongly suggests to us that f(a,b,c)=(a+b+c)×g(a,b,c) f(a, b, c) = (a+b+c) \times g(a, b, c) .

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

Putting this all together, this strongly suggests that

(a+b+c)(abc)(bca)(cab)f(a,b,c) (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)(abc)(bca)(cab) f(a, b, c) = - (a+b+c)(a-b-c)(b-c-a)(c-a-b)

Calvin Lin Staff - 6 years ago

Log in to reply

The equation 2(a4+b4+c4)=(a2+b2+c2)22(a^4 + b^4 + c^4) = (a^2 + b^2 + c^2)^2 is equivalent to a4b4c4+2a2b2+2a2c2+2b2c2=0.-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)(ab+c)(a+bc)=0,(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: K2=s(sa)(sb)(sc)=116(a+b+c)(a+b+c)(ab+c)(a+bc)=116(a4b4c4+2a2b2+2a2c2+2b2c2). \begin{aligned} 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{aligned}

Jon Haussmann - 6 years ago

Log in to reply

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

Calvin Lin Staff - 6 years ago

Log in to reply

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

Nihar Mahajan - 6 years ago

Log in to reply

Use the convention of S1=a+b+cS_1 = a+b+c, S2=ab+bc+caS_2=ab+bc+ca and S3=abcS_3 = abc, and Pn=an+bn+cnP_n = a^n+b^n+c^n. Then by Newton's Sum method:

P2=S1P12S2=02S2=2S2P_2 = S_1 P_1 - 2S_2 = 0 - 2S_2 = -2S_2

S4=S1P3S2P2+S3P1=0S2(2S2)+0=2S22S_4 = S_1P_3-S_2P_2+S_3P_1 = 0 -S_2(-2S_2) + 0 = 2S_2^2

Now LHS=2P4=4S22LHS = 2P_4 = 4S_2^2 and RHS=P22=(2S2)2=4S22=LHSRHS = P_2^2 = (-2S_2)^2 = 4S_2^2 = LHS

Chew-Seong Cheong - 6 years ago

Log in to reply

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

Mehul Arora - 6 years ago

Log in to reply

That's the most special case. -_-

Nihar Mahajan - 6 years ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...