Question

Let $a,b,c$ and $d$ be any four real numbers. Then, $a^n+b^n=c^n+d^n$ holds for any natural number $n$, if

(This question has some ambiguity, but appeared in WBJEE 2015 exam).

• A. $a+b=c+d$
• B. $a-b=c-d$
• C. $a+b=c+d,a^2+b^2=c^2+d^2$
• D. $a-b=c-d,a^2-b^2=c^2-d^2$

Answer 1

Answer: (D)

$a-b=c-d,a^2-b^2=c^2-d^2$

Option A

$a+b=c+d$ ....Given

Squaring on both sides, we get

$(a+b{)}^2=(c+d{)}^2$

$a^2+b^2+2ab=c^2+d^2+2cd$

$\therefore a^n+b^n\ne c^n+d^n$

Option B

$a-b=c-d$ ...Given

Squaring on both sides, we get

$(a-b{)}^2=(c-d{)}^2$

$a^2+b^2-2ab=c^2+d^2-2cd$

$\therefore a^n+b^n\ne c^n+d^n$

Option C

$a+b=c+d$ ...(i)

$a^2+b^2=c^2+d^2$ ....(ii)

Consider $(a+b{)}^2=(c+d{)}^2$

$a^2+b^2+2ab=c^2+d^2+2cd$

$\Rightarrow 2ab=2cd$ ...from (ii)

$\Rightarrow ab=cd$

So, $a^n+b^n\ne c^n+d^n$ for all $nϵN$.

Option D

$a-b=c-d$ ...(i)

$a^2-b^2=c^2-d^2$ ....(ii)

Consider $a^2-b^2=c^2-d^2$

$\Rightarrow (a-b)(a+b)=(c-d)(c+d)$

$\Rightarrow a+b=c+d$ ...from (i) ....(iii)

Adding eq(i) and (ii), we get

$2a=2c$

$\Rightarrow a=c$

$\therefore b=d$ ...from (iii)

So, $a^n+b^n=c^n+d^n$ for all $nϵN$.