New user? Sign up

Existing user? Sign in

Prove that there are no triples \((a,b,c)\) of positive integers satisfying \(2^a - 5^b7^c = 1 \).

Note by Paramjit Singh 3 years, 11 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

Suppose there exists a triple \( (a,b,c,) \)

Then \( 2^a \equiv 1 \pmod5 \)

Since \( ord(5) = 4\) we have \( a = 4a'\) for some \( a' \).

Rearranging, \(2^{4a'} - 1 = 5^b7^c\)

\( 2^{4a'} - 1 \equiv (2^2)^{2a'} - 1 \equiv 1^{2a'} - 1 \equiv 0 \pmod3 \)

But \(5^b7^c \not\equiv 0 \pmod3 \)

Therefore there exist no triples \( (a,b,c) \) which satisfy the given statement

Log in to reply

Nice.

Comment deleted Feb 15, 2014

\( 2^3 - 2^2 = 4 \equiv 1 \pmod {3} \). You need to rethink your solution

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestSuppose there exists a triple \( (a,b,c,) \)

Then \( 2^a \equiv 1 \pmod5 \)

Since \( ord(5) = 4\) we have \( a = 4a'\) for some \( a' \).

Rearranging, \(2^{4a'} - 1 = 5^b7^c\)

\( 2^{4a'} - 1 \equiv (2^2)^{2a'} - 1 \equiv 1^{2a'} - 1 \equiv 0 \pmod3 \)

But \(5^b7^c \not\equiv 0 \pmod3 \)

Therefore there exist no triples \( (a,b,c) \) which satisfy the given statement

Log in to reply

Nice.

Log in to reply

Comment deleted Feb 15, 2014

Log in to reply

\( 2^3 - 2^2 = 4 \equiv 1 \pmod {3} \). You need to rethink your solution

Log in to reply