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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: $$ a^n + b^n + c^n = d^n + e^n + f^n $$

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.

Shukant Pal
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  • Do such real numbers actually exist by the way? – Mr Pie May 13 '18 at 05:35
  • I don't know. I had great deal of problem in solving it - even no results on the Internet related to this were found. So I thought I could help others! – Shukant Pal May 13 '18 at 05:39
  • Does the book cited deal with symmetric polynomials and related topics? If so, it seems that a generalization may be proved easyly: Thm. Let $\sum_{i=1}^m a_i^k=\sum_{i=1}^m b_i^k$ for $1\leq k\leq m$. Then $\sum_{i=1}^m a_i^k=\sum_{i=1}^m b_i^k$ for all $k$. – Jens Schwaiger May 13 '18 at 07:46
  • @user477343: Following the discussions below it becomes clear that ${a,b,c}$ and also ${d,e,f}$is the set of roots of a polynomial of degree 3 common to both sets. Thus $d,e,f$ results from $a,b,c$ by applying a permutation. – Jens Schwaiger May 13 '18 at 08:02
  • @user477343 Since each triple is the set of three solutions of the same cubic polynomial equation in one variable, take any set of three real numbers. The second set is a permutation of the first. – Mark Bennet May 13 '18 at 08:02
  • This is for students in 11th class. Not for the advanced you are talking about. Don't know anything you just said. – Shukant Pal May 13 '18 at 10:05
  • @Sukant Kumar: Thats the reason why I asked what the book cited says about this topic. – Jens Schwaiger May 13 '18 at 14:45

2 Answers2

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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.

Mark Bennet
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  • $s, u, v$ are the first, second and third elementary symmetric polynomials in $a,b,c$. They may be expressed, as you have done, in terms of the first thre power sums $a^i+b^i+c^i$, $i=1,2,3$. The Main Theorem on elementary symmetric polynomials says that any symmetric polynomial in $a,b,c$ is a polynomial in the elementary symmetric polynomials. Thus $a^n+b^n+c^n=P_n(s,u,v)=d^n+e^n+f^n$. – Jens Schwaiger May 13 '18 at 07:41
  • @JensSchwaiger Of course. And I obtained the equation for $v$ by adding $0=p(a)+p(b)+p(c)=(a^3+b^3+c^3)-s(a^2+b^2+c^2)+u(a+b+c)-3v$ which is a relatively clean way to build such expressions. One gets a recurrence relation for the sums of powers. – Mark Bennet May 13 '18 at 08:06
  • But this doesn't use induction, right? – Shukant Pal Jul 15 '18 at 09:07
  • @SukantKumar Induction has two parts - the base case, and a step from $n$ to $n+1$. This shows a method for doing the second part by expressing sums of powers in terms of sums of lower powers. You need something like this to make the induction work. – Mark Bennet Jul 15 '18 at 09:13
  • How did you conclude $p(a) = p(b) = p(c)$ – tryst with freedom Feb 26 '21 at 17:24
  • @Buraian They are all zero, because $p(x)$ is constructed to have roots $a,b,c$. – Mark Bennet Feb 26 '21 at 17:29
  • Hmm fair but how did you conclude that 'd' must be also be a root? It is a cubic afterall @MarkBennet – tryst with freedom Feb 26 '21 at 17:30
  • @Buraian I created it that way using $(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$. The formulae for the coefficients in terms of the roots are known (for equations of all degrees) as Vieta's relations. – Mark Bennet Feb 26 '21 at 17:46
  • If $ d \neq a ,b,c$ then that means you have cubic with thre eroots, that is what is confusing me @MarkBennet – tryst with freedom Feb 26 '21 at 17:54
  • @Buraian Indeed - and the answer is: that the two sets are permutations of each other since each is a complete set of roots of the same cubic, and the three given equations are sufficient to establish that. However, it is not necessary to prove that to establish the induction required by the question. – Mark Bennet Feb 26 '21 at 19:53
  • Hmm then your identity holds for only permutations of (a,b,c) . I am not sure if I missed osmething, but in OP's question, he didn't put any relation between the variables on the left and the ones on the right [ a,b,c and d,e,f] @MarkBennet and also thanks for reading through my mistake lol [ I had meant a cubic with four roots but wrote three instead] – tryst with freedom Feb 26 '21 at 19:56
  • @Buraian Each of the two sets fo three maps to a cubic with those three roots. They both may to the same cubic, hence the sets are the same. Given the equality of sums of arbitrary powers this can also be shown by considering the one with the largest absolute value (or if two are the same ...). It is a consequence of what is given. – Mark Bennet Feb 26 '21 at 19:58
  • Let us continue this discussion in chat. – tryst with freedom Feb 26 '21 at 20:03
  • @Buraian Nothing to continue, I think. The two sets can be shown to be the same (which proves the proposition about powers). One can prove the proposition about powers by induction without showing the sets are the same. One can show the sets of the same from the fact that all the sums of powers are the same. It is just necessary to be clear what is being shown on the basis of which assumptions. – Mark Bennet Feb 26 '21 at 21:03
  • I think I'm starting to get it , could you suggest any 'soft' introductions to similar results about symmetric polynomials? @MarkBennet – tryst with freedom Feb 26 '21 at 21:08
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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,

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

Shukant Pal
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