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
$2 a=2 c \Rightarrow a=c$
$\Rightarrow b=d$ [using Eq. (iii)]
Thus, $a^{n}+b^{n}=c^{n}+d^{n}$ for all $n \in N$.