Prove that x^ 2n-1 +y^ 2n-1 is divisible by x + y for all n .

Question

Prove that $\text{x}^{2\text{n}-1}+\text{y}^{2\text{n}-1}$ is divisible by x + y for all $\text{n}\in\text{N}.$

✓

Answer

Let P(n): $\text{x}^{2\text{n}-1}+\text{y}^{2\text{n}-1}$ is divisible by (x + y)
For n = 1
$\text{x}^{2(\text{1})-1}+\text{y}^{2(\text{1})-1}$
= x + y
⇒ p(n) is true for n = 1
Let p(n) is true for n = k,
$\text{x}^{2\text{k}-1}+\text{y}^{2\text{k}-1}$ is divisible by (x + y)
$\text{x}^{2\text{k}-1}+\text{y}^{2\text{k}-1}=(\text{x + y})\lambda \ ...(1)$
We have to show that,
$\text{x}^{2\text{k}+1}+\text{y}^{2\text{k}+ 1}=(\text{x + y})\mu$
Now,
$\text{x}^{2\text{k}+1}+\text{y}^{2\text{k}+1}$
$\text{x}^{2\text{k}-1}.\text{x}^2+\text{y}^{2\text{k}+1}.\text{y}^2$
$=\Big[(\text{x}+\text{y})\lambda-\text{y}^{2\text{k}-1}\Big]\text{x}^2+\text{y}^{2\text{k}-1}.\text{y}^2$
$=(\text{x}+\text{y})\lambda\text{x}^2-\text{y}^{2\text{k}-1}.\text{x}^2+\text{y}^{2\text{k}-1}.\text{y}^2$
$=(\text{x}+\text{y})\lambda\text{x}^2-\text{y}^{2\text{k}-1}\big(\text{x}^2+\text{y}^2\big)$
$=(\text{x}+\text{y})\lambda\text{x}^2-\text{y}^{2\text{k}-1}\big(\text{x}+\text{y})(\text{x}+\text{y}\big)$
$=(\text{x}+\text{y})\Big[\lambda\text{x}^2-\text{y}^{2\text{k}-1}\big(\text{x}-\text{y}\big)\Big]$
$=(\text{x}+\text{y})\mu$
⇒ P(n) is true for n = k + 1
⇒ P(n) is true for $\text{n}\in\text{N}$ by PMI

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