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- Feb 14, 2012

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\(\displaystyle a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2\).

Prove that

\(\displaystyle a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4\).

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- Thread starter
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- #1

- Feb 14, 2012

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\(\displaystyle a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2\).

Prove that

\(\displaystyle a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4\).

Square and Simplify both sides. You should get \(\displaystyle 4\cdot(a^{2}-ab+b^{2})^{2} = 4\cdot(c^{2}-cd+d^{2})^{2}\)

Divide that equation by -2. You'll just have to trust me on this.

Subtract the quartic equation above from the given quartic equation. See if anything looks familiar in the result.

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Square and Simplify both sides. You should get \(\displaystyle 4\cdot(a^{2}-ab+b^{2})^{2} = 4\cdot(c^{2}-cd+d^{2})^{2}\)

Divide that equation by -2. You'll just have to trust me on this.

Subtract the quartic equation above from the given quartic equation. See if anything looks familiar in the result.

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Now we consider the expression \(\displaystyle a^2+b^2+(a-b)^2\), if we square it we will end up with:

\(\displaystyle (a^2+b^2+(a-b)^2)^2=a^4+b^4+(a-b)^4+2(a^2+b^2)^2+2a^2b^2-4ab(a^2+b^2)\).

Since \(\displaystyle a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2\), we will also have:

\(\displaystyle (a^2+b^2+(a-b)^2)^2=c^4+d^4+(c-d)^4+2(c^2+d^2)^2+2c^2d^2-4cd(c^2+d^2)\).

Hence, it is obvious that if we could prove

\(\displaystyle 2(a^2+b^2)^2+2a^2b^2-4ab(a^2+b^2)=2(c^2+d^2)^2+2c^2d^2-4cd(c^2+d^2)\) (*)

then we can conclude that

\(\displaystyle a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4\) must hold.

From \(\displaystyle a^2+b^2-ab=c^2+d^2-cd\), we get

\(\displaystyle a^2+b^2=c^2+d^2-cd+ab\) (**)

If we substitute (**) into only the left side of (*), we see that

\(\displaystyle 2(a^2+b^2)^2+2a^2b^2-4ab(a^2+b^2)\)

\(\displaystyle =2(c^2+d^2-cd+ab)^2+2a^2b^2-4ab(c^2+d^2-cd+ab)\)

\(\displaystyle =2(c^2+d^2-cd)^2+4ab(c^2+d^2-cd)+2a^2b^2+2a^2b^2-4ab(c^2+d^2-cd)-4a^2b^2\)

\(\displaystyle =2(c^2+d^2-cd)^2\)

\(\displaystyle =2(c^2+d^2)^2+2c^2d^2-4cd(c^2+d^2)+2c^2d^2\)

Therefore, we can deduce that \(\displaystyle a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4\) must be true since we're given \(\displaystyle a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2\).

- Jan 25, 2013

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let $a^2+b^2=x , \,\, c^2+d^2=y$Let \(\displaystyle a\), \(\displaystyle b\), \(\displaystyle c\) and \(\displaystyle d\) be positive real numbers such that

\(\displaystyle a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2-------(1)\).

Prove that

\(\displaystyle a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4\).

from (1) we have x-ab=y-cd

then 2$(x-ab)^2$=2$(y-cd)^2-----(2)$

expansion left side of (2):

2$x^2-4abx+2a^2b^2=2(a^4+2a^2b^2+b^4)-4ab(a^2+b^2)+2a^2b^2$

$=a^4+b^4+a^4-4a^3b+6a^2b^2-4ab^3+b^4=a^4+b^4+(a-b)^4$

likewise expansion the right side of (2):

we get $c^4+d^4+(c-d)^4$

and the proof is done !

\(\displaystyle a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4\)