Question

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

• 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)

From option (d), we have
$a-b=c-d$ ...(i)
and $a^2-b^2=c^2-d^2$ ...(ii)
Consider, $a^2-b^2=c^2-d^2$
$\Rightarrow (a+b)(a-b)=(c-d)(c+d)$
$\Rightarrow a+b=c+d$ ...(iii) [using Eq. (i)]
On adding Eqs. (i) and (iii), we get
$2a=2c\Rightarrow a=c$
$\Rightarrow b=d$ [using Eq. (iii)]
Thus, $a^n+b^n=c^n+d^n$ for all $n\in N$.