21. $x^{2 x} - y^{2 x} is divisible by x + y$

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Payal Gupta | Contributor-Level 10

9 months ago

21. Let $\begin{array}{l} P (n) : x^{2 n} - y^{2 n} is divisible by x + y \end{array}$

$Putting x = 1,$

$P (1) = x^{2} - y^{2} is divisible by x + y$ or

$(x + y) (x - y)isdivisibleby x + y,$ which is true.

Assume that P(k) is true for some natural no. k

$P (k) = x^{2 k} - y^{2 k}$ is divisible by x + y

i.e. $x^{2 k} - y^{2 k} = a (x + y)$ where $\in z$

$\Rightarrow x^{2 k} = a (x + y) + y^{2 k}$ (1)

Now, let us prove P(k +1) is true.

$P (k + 1) : x^{2 (k + 1)} - y^{2 (k + 1)}$

$= x^{2 x + 2} - y^{2 x + 2}$

$= x^{2} \cdot x^{2 k} - y^{2} \cdot y^{2 k}$

$= x^{2} [a (x + y) + y^{2 k}] - y^{2} \cdot y^{2 k} [using(1)$
$]$

$= a x^{2} (x + y) + x^{2} \cdot y^{2 k} - y^{2} \cdot y^{2 k}$

$= a x^{2} (x + y) + y^{2 k} (x^{2} - y^{2})$

$= a x^{2} (x + y) + y^{2 k} (x + y) (x - y)$

$= (x + y) [a x^{2} + y^{2 k} (x - y)]$

$= b (x + y)where b = [a x^{2} + y^{2 k} (x - y)] \in z$

$\Rightarrow x^{2 k + 2} - y^{2 k + 2} is divisible by x + y$

$∴$ P(k+ 1) is true where P (k) is true.

Hence, by P.M.I. P(n) is true for all natural number i.e.

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21. x2x−y2x is divisible by x+y