Divisibility example.

If $d|n$ then prove that ${ 2 }^{ d }-1|{ 2 }^{ n }-1$.

This is my solution

If $d|n$ then $n=dk$ for some $k$ .Then we have to prove ${ 2 }^{ d }-1|{ 2 }^{ dn }-1$ or ${ 2 }^{ d }-1|{ ({ 2 }^{ d }) }^{ k }-1$ .Now let $x={ 2 }^{ d }$ therefore we have to prove $x-1|{ x }^{ k }-1$ which is true by Remainder theorem.

#NumberTheory #Remainder-FactorTheorem #Proofs #Divisibility #Proof

Easy Math Editor

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`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`	$\sin \theta$
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Comments

Your proof is awesome.

Adarsh Kumar - 6 years, 6 months ago

nice....

vivek waghmare - 6 years, 6 months ago

Thanx.@Adarsh Kumar

shivamani patil - 6 years, 6 months ago

NICE but too EASY.

Krishna Ar - 6 years, 6 months ago

Math is neither easy nor difficult for her lover.@Krishna Ar

shivamani patil - 6 years, 6 months ago

haha

Krishna Ar - 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}`	$\boxed{123}$