For what values of m in mathbb{N},does the di | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We want to find $m \in \mathbb{N}$ such that $(x+y)$ divides $(x^m+y^m)$.Let $P(m)$ be the statement that $(x+y) \mid (x^m+y^m)$.For $m=1$: $x^1+y

Vedclass · Accepted Answer

Odd numbers

Vedclass · Answer

(B) We want to find $m \in \mathbb{N}$ such that $(x+y)$ divides $(x^m+y^m)$.Let $P(m)$ be the statement that $(x+y) \mid (x^m+y^m)$.For $m=1$: $x^1+y^1 = x+y$,which is clearly divisible by $x+y$. Thus,$P(1)$ is true.For $m=2$: $x^2+y^2$ is not generally divisible by $x+y$ (e.g.,if $x=1, y=1$,$2$ is divisible by $2$,but if $x=2, y=1$,$5$ is not divisible by $3$). Thus,$P(2)$ is false.For $m=3$: $x^3+y^3 = (x+y)(x^2-xy+y^2)$,which is divisible by $x+y$. Thus,$P(3)$ is true.In general,$x^m+y^m$ is divisible by $x+y$ if and only if $m$ is an odd natural number.Therefore,the divisibility holds for all odd numbers $m \in \mathbb{N}$.

For what values of $m \in \mathbb{N}$,does the divisibility $(x+y) \mid (x^m+y^m)$ hold?

Similar Questions

The last digit of $(3^P + 2)$ is,where $P = 3^{4n}$ and $n \in N$.

The remainder when $3^{91}$ is divided by $80$ is:

The remainder when $3^{2022}$ is divided by $5$ is

For a positive integer $n$,$n^{3} + 2n$ is always divisible by

If the expression $5^{2n} - 48n + k$ is divisible by $24$ for all $n \in N$,then the least positive integral value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module