ISI 2017 PCB A2

Question

Let $a,b,c$ and $d$ be real number such that $a+b=c+d$ and $ab=cd$. Prove that $a^{n}+b^{n}=c^{n}+d^{n}$ for all positive integer $n$.

Comments

Do it with the help of induction on n (strong form)

at the time of proving for n= k+1 use

(a^k +b^k) (a+b) = (c^k +d^k) (c+d)

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No I don't think it will help it won't be equal to a$^{n+1} + b^{n+1}$

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My previous comment was a HINT. Whole solution is done in the answer section. Please read it.

Answer 1

Statement $P(n) : a^n +b^n = c^n +d^n $

here we have to prove $P(n)$ is true for all $n \in N$

So, $P(1) : a+b = c+d$

hence $P(1)$ is true (given in the question)

Now, assume  $P(m) : a^m +b^m = c^m + d^m $ is true for all $m \leq k$ for some $m,n \in N$.................. (1)

Now, with the help of the assumption (1)  we have to prove $a^{k+1} + b^{k+1}= c^{k+1} + d^{k+1}$

Now, 

        $(a^k + b^k) (a+b) = (c^k +d^k) \ (c+d)$  ( since we have assumed in $1$ that $a^m +b^m = c^m+d^m$ for all $m \leq k$ )

$\Rightarrow$     $a^{k+1} + ab^k + a^k b + b^{k+1} = c^{k+1} + cd^k + c^k d + d^{k+1}$

$\Rightarrow$     $a^{k+1}+ b^{k+1} +ab( b^{k-1} + a^{k-1}) = c^{k+1} + d^{k+1} + cd(d^{k-1} + c^{k-1})$

$\Rightarrow$     $a^{k+1}+ b^{k+1} = c^{k+1} +d^{k+1}$  ( Here $ab = cd$ (given) and $(a^{k-1} + b^{k-1}) = (c^{k-1} +d^{k-1})$ (from assumption (1))

So, $P(n)$ is true for all $n \in N$

Comments

such a beautiful solution..thanku!