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If $a+b=c+d$, and $a^3+b^3=c^3+d^3$, then prove that: $$a^{2011}+b^{2011} = c^{2011}+d^{2011}$$

Approach:

From $$a+b=c+d$$ We have: $$(a+b)^2 = (c+d)^2$$ $$a^2+b^2+2ab = c^2+d^2+2cd .... (1)$$ Similarly, we have: $$a^2+b^2-ab = c^2+d^2-cd ....(2)$$ from $a^3+b^3=c^3+d^3$

$(1)-(2)$, gives us $ab=cd$

I wanted to try the below formula explicitly for $n=$ odd, but it isn't working out well.

$$a^n+b^n= (a+b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2-a^{n-4}b^3+.....+a^2b^{n-3}-ab^{n-2}+b^{n-1})$$

Where $n$ is odd.

I got $a-b=c-d$ by using the above information, but I am unable to take $a-b$ common due to the last term as follows:

$$a^n+b^n= (a+b)\ \ \ [ \ \ a^{n-2}(a-b) + a^{n-4}b^2(a-b)+ a^{n-6}b^4(a-b).....+ab^{n-3}(a-b)+b^{n-1}]$$

I want to generalize the result for all odd numbers, but I'm unable to do so. Any help or hints would be most appreciated! Thanks.

saulspatz
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2 Answers2

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Show that $(a-b)^2 = (c-d)^2$. Then consider two cases.

Case 1: $$a-b = c-d.$$ We also have $a+b= c+d$. This gives us $a=c$ and $b=d$, and we're done.

Case 2: $$a-b = -(c-d).$$ We're given $a+b =c+d$. Then $a=d$ and $b=c$. Again, we're done.

Matha Mota
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Alternatively:

Let $x_n=a^{2n+1}+b^{2n+1}=a(a^2)^n+b(b^2)^n$. Then $x_{n+2}=(a^2+b^2)x_{n+1}-(a^2b^2)x_n$ and so $x_n$ is determined by $x_0$ and $x_1$.

Let $y_n=c^{2n+1}+d^{2n+1}$. Then $y_{n+2}=(c^2+d^2)y_{n+1}-(c^2d^2)y_n$ and so $y_n$ is determined by $y_0=x_0$ and $y_1=x_1$.

We only have to prove that $a^2+b^2=c^2+d^2$ and $a^2b^2=c^2d^2$, which you have done.

lhf
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