Equation without a proof

(2n±1)2+⌊(2n±1)22⌋2=⌈(2n±1)22⌉2{ (2n\pm 1) }^{ 2 }+{ \left\lfloor \frac { { (2n\pm 1) }^{ 2 } }{ 2 } \right\rfloor }^{ 2 }={ \left\lceil \frac { { (2n\pm 1) }^{ 2 } }{ 2 } \right\rceil }^{ 2 }

I somehow found an equation that can generate Pythagorean triplets as shown above for any integral value of nn, regardless of positive or negative.

But I can't find a formal proof... So any approach or suggestions?

#NumberTheory

Note by William Isoroku
5 years, 1 month 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} ai−1 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 sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Let ⌊(2n±1)22⌋=v\left \lfloor \frac{(2n \pm 1)^2}{2} \right \rfloor = v Then, ⌈(2n±1)22⌉=v+1(2n±1)2=2v+1\left \lceil \frac{(2n \pm 1)^2}{2} \right \rceil= v+1\\ (2n \pm 1)^2= 2v+1

Hence, the conjectured equation is equivalent to (2v+1)+(v)2=(v+1)2(2v+1)+(v)^2=(v+1)^2 Which is obviously true! Q.E.D.\large Q. E. D.

A Former Brilliant Member - 5 years ago

Look at the 2n+1 case. (2n+1)^2=4n^2+4n+1. As in the expression, we are dividing by 2. The 1/2 will be canceled when the floor is taken, so we will have (2n^2+2n)^2 for that part. The LHS will then become (2n+1)^2+(2n^2+2n)^2. Now we have the RHS. Applying a similar method we get 4n^2+4n+1 to round up with the ceiling, so now we have (2n^2+2n+1)^2. Now equate the LHS and RHS, yielding (2n+1)^2+(2n^2+2n)^2=(2n^2+2n+1)^2. Dy difference of square we know this is true because the (2n+1)^2 must equal (2n^2+2n+2n^2+2n+1). Hence proven. The other case simplifies in a nearly identical way. If you need more specifications, just ask. Nice formula.

Sal Gard - 5 years ago
×

Problem Loading...

Note Loading...

Set Loading...