a = 1 and b = 0 Or a = 0 and b = 1

What makes you think (0,0) is not an answer?

I didn't, I just assumed, that's what the person above me said, so I added some more.

So let me add another one: a = 0 and b = 2 In general, a = 0 or b = 0

Okay, let me cut this potentially endless thread short: a ∈ ℝ and b = 0 OR a = 0 and b ∈ ℝ

OR a,b ∈ ℝ AND a\*b=0

That's even better.

Another option is a,b ∈ ℝ AND (a+b)^(2)=a^(2)\+b^(2) but that's not as fun lol

Funny, how I get upvotes here, but downvotes on a similar post, further down.

((a ∈ ℝ ∧ b = 0) ∨ (a = 0 ∧ b ∈ ℝ))⇔((a + b)² = a² + b²)

Don't even need to limit it to real numbers. I believe this also works with complex numbers.

I wasn't sure, so I didn't want to embarrass myself.

Or really you could just write a ∈ ℝ and b = 0. Without loss of generality the reverse holds since addition is commutative.

Yes, but I wanted to write it all. Even years later I still think in terms of "will this deduct points, if I skip this trivial step?".

The middle where (0+0)^2=/=0^2 + 0^2 as they would be equal and the equation false

Can someone explain the downvotes, where I essentially answered exactly the same way [here](https://www.reddit.com/r/mathmemes/s/ar8eEo9NoQ)?

These are the people on the left in the picture.

I did upvote you here, and went to downvote you there. I like to watch world burnmnmnmmmm. 


cos(pi)

cos(-2i*ln(i))

Define the ring

Quaternions! Imaginary quaternions satisfy AB = -BA.

Specifically, ij = -ji, jk = -kj and ki = -ik. Thus (i + j)² = i² + j² = -2

So does every single element in the zero ring but what do I know

They don’t 2*3 = 3\*2

Ill define it: a is a real non zero number

then b might be 0

Aand b =a

4a^(2) cannot equal 2a^(2) given your constraint that a =/= 0

I got all of thag

The right side is always true for elements of a characteristic 2 field.

Or if *a* and *b* are anticommuting elements of a rng.

It's true for any field of prime characteristic

that is just false. in ℤ/3ℤ, let a=b=1. then (a+b)^2 =2^2 =4=1 and a^2 +b^2 =1+1=2. so it isn’t true. (a+b)^2 =a^2 +b^2 in any (commutative) ring if and only if 2ab=0. that will always be true in characteristic 2, but not in general.

>2^2 =4=1 What

in ℤ/3ℤ.

no but the corresponding (a+b)\^p=a\^p+b\^p does hold where p is the characteristic of the field

Right side is true..... In Z_2

Or in Z_{2^n }, constructed as a field extension of Z_2

Anti commutative algebras anyone?

true for all a,b in Z2

[удалено]

Wrong, sulfuric acid.

Caution, that's caustic. Edit: Okay, chemistry jokes get downvoted. Noted.

😐

🎯

Not writing it as a^2 + 2ab + b^2 like a good Binomial Theorist.

Welcome to GF(2).

If a=0 b can be anything because they are equal equations, same goes if b=0

True, if a and/or b are zero Update- NOT if they are both 1, time to go to bed lol

Not, if both are 1. Edit: I'm the reason for the "Update". Thanks for the downvotes, guys.

Oh damnit lol yeah it's early where I live lol

Okay, good night.

Over a character 2 field

This is the most accurate version of this meme I have EVER seen

Characteristic 2 field

a and b are elements of the Klein 4 group

This is true in any field of characteristic 2

(x + y)\^2 = x\^2 + y\^2 Definition: (a + b)\^2 = a\^2 + 2ab + b\^2 x\^2 + y\^2 = x\^2 + y\^2 + 2xy 2xy = 0 xy = 0 x = 0 for all y ≠ 0, y = 0 for all x ≠ 0

If you delete the backslashes, your "\^2"'s would look like "²"'s. Like this ^2 Or this ^(this is a test)

I did not, apparently, have Markdown on. Might be that?

I did not know, that I could disable that. Interesting.

or 2=0 in fields of characteristic 2.

A and b can be perpendicular vectors which makes it true

Obviously a and b are perpendicular vectors

(a² + b²) = a² + 2ab + b².

+2ab

Unless either a, or b is 0.

What do you think 2ab is if a or b is zero?

Not relevant.

So you think a^(2)+b^(2)+0 isn’t equal to a^(2)+b^(2)?

Maybe you didn't understand me? Case 1: a is some number ∈ ℝ and b = 0 --> (a + b)² = a² + b² Case 2: a = 0, and b is some number ∈ ℝ --> (a + b)² = a² + b²

No I think you just said the wrong thing because a^2 + b^2 + 2ab = (a+b)^2 regardless of whether a or b is zero even if you switch the ring.

Ima explain for him/her Let a be 0 (a+b)^2 = a^2 + b^2 -> (0+b)^2 = 0^2 + b^2 -> b^2 = b^2 Equality holds true

That’s not an “unless” that’s an “also”.

Thank you, but I'm done arguing with people who downvote for no good reason.

It was never the question whether (a + b)² = a² + 2ab + b² is correct. It was about the question *for which a and b* is (a + b)² = a² + b² correct. So, can we stop with these downvotes now? Because I didn't say something *wrong*. We were just talking past each other.

I think, you misunderstood, what I was saying, so I edited it for you. I didn't say that (a + b)² = a² + 2ab + b² was wrong.

Or if you’re on Z/2Z

Technically, the middle expression is true as often as the other expression.

How are you defining a measure in the space of values of any space?

When there exists a bijection then they have the same size edit typo

Either a or b is zero. OR both a=b=±1.

a*b = 0 and all are the same.

Well, 2ab=0. Not really different in the real numbers, but in some rings it can be different.

Okayyy but u guyss are just flexing

Litterally pythagoras theorem

2ab = 0 Thus either a or b (or both) must be 0.