Prove that $ a^n + b^n + c^n = d^n + e^n + f^n $ by induction

Question

If $a,b,c,d,e,f$ are six real numbers such that: $$ a + b + c = d + e + f $$ $$ a^2 + b^2 + c^2 = d^2 + e^2 + f^2 $$ $$ a^3 + b^3 + c^3 = d^3 + e^3 + f^3 $$

Prove by mathematical induction that: $$ a^n + b^n + c^n = d^n + e^n + f^n $$

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I tried solving this question by correlating to $$ a^k + b^k = (a + b)(a^{k-1} + b^{k-1}) - ab(a^{k-2} + b^{k-2}) $$

The problem is that terms with $abc$ do not come in common while expanding. Could you please help me solve this question?

This question originates from the level III excercise of SK Goyal's book of Algebra for JEE Mains and Advanced.

Answer 1

Consider the polynomial $p(x)=x^3-sx^2+ux-v$ where $s=a+b+c=d+e+f$, $u=ab+bc+ca=\frac 12\left((a+b+c)^2-a^2+b^2+c^2\right)=de+ef+fd$ and $v=abc=\frac 13\left((a^3+b^3+c^3)-s(a^2+b^2+c^2)+u(a+b+c)\right)=def$

Then $p(a)=p(b)=p(c)=p(d)=p(e)=p(f)=0$ and you can use $$a^rp(a)+b^rp(b)+c^rp(c)=0$$ to obtain an expression for the sum $a^{r+3}+b^{r+3}+c^{r+3}$ in terms of sums of lower powers and the common constants $s,u,v$. This can be used for the inductive step.

Answer 2

Let $ P(n): a^n + b^n + c^n = d^n + e^n + f^n $,

Step I:

For $ n=1,2,3 $ it is already given that $P(n)$ is true. There is no need to do extra cross-checking here.

Step II:

Assume that for $ n=k,k-1,k-2 $ the result is true, i.e.

$$ P(k): a^{k} + b^{k} + c^{k} = d^{k} + e^{k} + f^{k} $$ $$ P(k-1): a^{k-1} + b^{k-1} + c^{k-1} = d^{k-1} + e^{k-1} + f^{k-1} $$ $$ P(k-2): a^{k-2} + b^{k-2} + c^{k-2} = d^{k-2} + e^{k-2} + f^{k-2} $$

Step III:

For $n=k+1$,

$$ P(k+1): a^{k+1} + b^{k+1} + c^{k+1} = d^{k+1} + e^{k+1} + f^{k+1} $$

We know that,

$$ a^{k+1} + b^{k+1} + c^{k+1} = (a + b + c)(a^k + b^k + c^k) - a(b^k + c^k) - b(c^k + a^k) - c(a^k + b^k) \tag{1} $$

$$ a^{k+1} + b^{k+1} + c^{k+1} = (a^2 + b^2 + c^2)(a^{k-1} + b^{k-1} + c^{k-1}) - a^2(b^{k-1} + c^{k-1}) - b^2(c^{k-1} + a^{k-1}) - c^2(a^{k-1} + b^{k-1}) \tag{2} $$

Adding $(1)$ and $(2)$, $$ 2(a^{k+1} + b^{k+1} + c^{k+1}) = (a+b+c)(a^k+b^k+c^k) + (a^2+b^2+c^2)(a^{k-1}+b^{k-1}+c^{k-1}) - a(b^k+c^k) - a^2(b^{k-1}+c^{k-1}) - b(c^k+a^k) - b^2(c^{k-1} + a^{k-1}) - c(a^k+b^k) - c^2(a^{k-1} + b^{k-1}) $$

Let's try to simply our equation here, $$ 2(a^{k+1} + b^{k+1} + c^{k+1}) = (a+b+c)(a^k+b^k+c^k) + (a^2+b^2+c^2)(a^{k-1}+b^{k-1}+c^{k-1}) - 2ab(a^{k-1}+b^{k-1}) - 2bc(b^{k-1}+c^{k-1}) - 2ca(c^{k-1}+a^{k-1}) $$

We've reached here, now as you said terms with $abc$ are hard to find, yes they are but we'll change the terms a bit here:

$$ 2(a^{k+1} + b^{k+1} + c^{k+1}) = (a+b+c)(a^k+b^k+c^k) + (a^2+b^2+c^2)(a^{k-1}+b^{k-1}+c^{k-1}) - 2ab(a^{k-1}+b^{k-1}+c^{k-1}) - 2bc(a^{k-1}+b^{k-1}+c^{k-1}) - 2ca(a^{k-1}+b^{k-1}+c^{k-1}) + 2(abc^{k-1} + a^{k-1}bc + ab^{k-1}c) $$

$$ 2(a^{k+1} + b^{k+1} + c^{k+1}) = (a+b+c)(a^k+b^k+c^k) + (a^2+b^2+c^2)(a^{k-1}+b^{k-1}+c^{k-1}) - 2(ab+bc+ca)(a^{k-1}+b^{k-1}+c^{k-1}) + 2abc(c^{k-2} + a^{k-2} + b^{k-2}) $$

Oh! Dang it, we've got $ab+bc+ca$ and $abc$ to make our work harder. But there is a simple solution to this:

Given that $a+b+c=d+e+f$, $a^2+b^2+c^2=d^2+e^2+f^2$, $a^3+b^3+c^3=d^3+e^3+f^3$,

Result I: $$ (a+b+c)^2 - (a^2+b^2+c^2) = (d+e+f)^2 - 2(de+ef+fd) $$ $$ 2(ab+bc+ca) = 2(de+ef+fd) \tag{3} $$

Result II:

$$(a+b+c)^3 - 3(a+b+c)(a^2+b^2+c^2) + 2(a^3+b^3+c^3) = (d+e+f)^3 - 3(d+e+f)(d^2+e^2+f^2) + 2(d^3+e^3+f^3)$$ $$ 6abc=6def $$ $$ abc = def \tag{4} $$

Putting in $(3)$,$(4)$,$P(k)$,$P(k-1)$,$P(k-2)$ in our equation, $$ 2(a^{k+1} + b^{k+1} + c^{k+1}) = (d+e+f)(d^k+e^k+f^k) + (d^2+e^2+f^2)(d^{k-1}+e^{k-1}+f^{k-1}) - 2(de+ef+fd)(d^{k-1}+e^{k-1}+f^{k-1}) + 2def(d^{k-2} + e ^{k-2} + f^{k-2}) $$

The way we have factored out $P(k+1)$ in terms of $a,b,c$ we could do that with $d,e,f$ so we don't have to again simplify the right-hand side,

$$ 2(a^{k+1} + b^{k+1} + c^{k+1}) = 2(d^{k+1} + e^{k+1} + f^{k+1}) $$ $$ \therefore a^{k+1} + b^{k+1} + c^{k+1} = d^{k+1} + e^{k+1} + f^{k+1}$$