Proof Contest Day 1

Prove that there no solutions for

5a+8b=c2 \large{5^a+8^b=c^2 }

For a,b,cZ+a,b,c \in \mathbb{Z^{+}} and bb is not an even positive integer.

#NumberTheory

Note by Department 8
5 years, 5 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. 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 1

paragraph 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 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

c20,1,4,9,6,5(mod10)c^2 \equiv 0, 1, 4, 9, 6, 5 \pmod{10}

5a5(mod10)5^a \equiv 5 \pmod{10}

8b8,2(mod10)(b is odd)8^b \equiv 8, 2 \pmod{10} (b \text{ is odd})

5a+8b3,7(mod10)5^a + 8^b \equiv 3,7 \pmod{10}

Clearly, the first statement and the last statement share no residues. Thus proven.

Sharky Kesa - 5 years, 5 months ago

Log in to reply

Haha, the same way I solved the question!

Department 8 - 5 years, 5 months ago

Log in to reply

Nice problem. By the way , from where have you collected these problems?

Priyanshu Mishra - 5 years, 5 months ago

Log in to reply

@Priyanshu Mishra Own thinking. Try other problems too from this set.

Department 8 - 5 years, 5 months ago

Log in to reply

@Department 8 Trying others.

Priyanshu Mishra - 5 years, 5 months ago

What's mod? And how are 6 and 5 also included?

Yuki Kuriyama - 5 years, 5 months ago

Log in to reply

ab(modc) a \equiv b \pmod c

Implies a leaves same remainder as b when divided by c.

Department 8 - 5 years, 5 months ago

yes it must have i;e solve by indicies a+b=2

amar nath - 5 years, 5 months ago

Log in to reply

The bases are not same. Additionally, there's no multiplication involved so you cannot add the indices.

Arulx Z - 5 years, 4 months ago
×

Problem Loading...

Note Loading...

Set Loading...