Proving equality

anemone

MHB POTW Director
Staff member
Let $$\displaystyle a$$, $$\displaystyle b$$, $$\displaystyle c$$ and $$\displaystyle d$$ be positive real numbers such that
$$\displaystyle a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$$.

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

tkhunny

Well-known member
MHB Math Helper

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.

MarkFL

Staff member

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.
Hello tkhunny,

Problems posted in this forum are meant as challenges to our members. The OP should already have a full and correct solution ready to post, and is looking to see if anyone else can solve it, and in the case that no one does after at least a week, will post their solution. If others do solve it, but use a different method, then the OP should post their solution as well so that we get multiple ways of attacking the problems.

So, unlike our other forums where we expect only hints and suggestions on how to proceed as given help, this forum is meant for full solutions to be posted, as stated in our guidelines:

edit: These guidelines were only recently posted, so it is understandable that not everyone has seen them yet.

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tkhunny

Well-known member
MHB Math Helper
Whoops. I didn't miss the updates, but I did manage not to check the forum title. I usually avoid the "Challenge" Forums. Not quite sure how I wandered in unawares. Thanks for the heads-up.

anemone

MHB POTW Director
Staff member
We're given that $$\displaystyle a^2+b^2+(a-b)^2=c^2+d^2+(c-d)^2$$.

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

Albert

Well-known member
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
$$\displaystyle a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4$$.
let $a^2+b^2=x , \,\, c^2+d^2=y$
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 !
$$\displaystyle a^4+b^4+(a-b)^4=c^4+d^4+(c-d)^4$$