# No triples? :(

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

Note by Paramjit Singh
4 years, 4 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> 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"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

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

- 4 years, 4 months ago

Nice.

- 4 years, 4 months ago

Comment deleted Feb 15, 2014

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

- 4 years, 4 months ago