Prove that $x^{n}$ + $y^{n}$ is divisible by x+y if n is odd

Can anybody prove it.Please.

#NumberTheory

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Hi,@Ayush Choubey If n is odd, then $x^{n}+y^{n}=(x+y)(x^{n-1}-x^{n-2}y+x^{n-3}y^{2}-....+y^{n-1})$ which can be proved easily by expanding or using induction.

Abhijeet Verma - 5 years, 9 months ago

Let's denote $p(x) = x^{n}+y^{n}$ . Due to remainder theorem (little Bezout's theorem) remainder will be $p(-y)= (-y) ^{n} + y^{n}=0$ , because n is odd. Therefore the remainder is 0, so p(x) is divisible by x+y.

Karan Pedja - 6 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Prove that xnx^{n}xn + yny^{n}yn is divisible by x+y if n is odd

Comments

Prove that $x^{n}$ + $y^{n}$ is divisible by x+y if n is odd