You should be able to solve this.

[Anonymous No.15107427]

For which if any pairs of integers [math]p,q \in Z[/math] is it possible to partition [math]Z[/math] into three subsets such that for each [math]n\in Z[/math] the integers [math]n, n+p, n+q[/math] are in different sets?

[Anonymous No.15107464]

>>15107427
p = 1 mod 3
q = 2 mod 3

[Anonymous No.15107623]

>>15107464

[Anonymous No.15107637]

>>15107623
Well, I could write a 1000-page book that 1+1=2 like Bertrand Russell, or I could assume that it's obvious enough

[Anonymous No.15107648]

>someone posts wrong answer with no argument and calls anyone who questions it retarded
every time

[Anonymous No.15107668]

>>15107623
proof is left as an exercise to the reader

[Anonymous No.15107670]

>>15107648
Hey now, I didn't say that was an exhaustive list

[Anonymous No.15107874]

>>15107427

Assume [math]p,q \geq 0[/math] otherwise it is easy to change it into an equivalent case.

The necessary and sufficient condition is that [math] 3 | \frac{p+q}{\gcd(p+q)} [/math], which is equivalent to [math] p = \pm k \bmod 3k [/math] and [math] p = \mp k \bmod 3k [/math] for some [math] k>0 [/math].

[math] x [/math] and [math] x + p + q [/math] are in the same set because they must be in different sets from both [math] x + q [/math] and [math] x + q [/math].

We can consider the problem in [math] \mathbb{Z}_{p+q} [/math] now. Now [math] x,x+p,x+2p [/math] are all in different sets and [math] x [/math] is in the same set as [math] x+3p [/math]. It is necessary that the subgroup generated by [math] p [/math] in [math] (\mathbb{Z}_{p+q},+) [/math] has size multiple of [math] 3 [/math] which gives us the necessary condition that [math] 3 | \frac{p+q}{\gcd(p+q)} [/math]. Construction is obvious.

[Anonymous No.15107887]

>>15107874

fuck i meant [math] 3 | \frac{p+q}{\gcd(p,q)} [/math]

[Anonymous No.15107890]

>>15107427
p = 1 mod 2
q = 2 mod 4

[Anonymous No.15107916]

You sould be able to slove this, at no point you are able to prof that your awareness is a web but not a flied. That means in the second case you get sucked into the main sequence Star, then into Arcturus and then into the galaxy core and squished to an 1 dimensional fractal.

the question is are you Jewish enough to flee into Boötes Void or not.

[Anonymous No.15107944]

Additive and combinatorial number theory question
[eqn]Z\cap Z+p=Z\cap Z+q=\emptyset[/eqn]
That is, [math]Z\cap Z+|p-q|=\emptyset[/math] and it happens that [math]|p-q|\notin Z-Z[/math] and besides [math]p, q\notin Z[/math] and also from the first equation in follows that [math]p, q\notin Z-Z[/math] and also one can find that [math](p, q)\notin (Z+a)^2[/math], finally we deduce that for any [math]a\in\mathbb{Z}[/math]
[eqn](p, q)\notin (Z+a)^2\cup(Z-Z)^2[/eqn]
But then notice that [math](Z-Z)\subseteq Z+a[/math] so we are left only with the restriction that
[eqn](p, q)\notin\bigcup_{z\in\mathbb{Z}}{(Z+z)^2}[/eqn]

[Anonymous No.15107950]

>>15107874
>x,x+p,x+2p are all in different sets
x,x+p and x+p,x+2p are in different sets but that doesn't imply x,x+2p are

> Construction is obvious.
>IT'S OBVIOUS!!

[Anonymous No.15107956]

>>15107944
Actually there's no need for absolute value since our domain is [math]\mathbb{Z}[/math]

[Anonymous No.15107966]

>>15107950

> x,x+p and x+p,x+2p are in different sets but that doesn't imply x,x+2p are
we are working in [math] \mathbb{Z}_{p+q} [/math] retard. [math] (x+p,x+2p,x+p+q) = (x+p,x+2p,x) [/math].

> IT'S OBVIOUS!!
literally put integer [math] x [/math] in the set [math] \lfloor \frac{x}{\gcd(p,q)} \rfloor \bmod 3 [/math]

[Anonymous No.15107968]

>>15107427
>>15107464
too much thinking, just decide it by a duel. And the winning factors of said duel should just be who lives and dies. Loser is weak, winner is strong. Pretty easy, don’t be a nerd

[Anonymous No.15107969]

>>15107966
See how much clearer your arguments become after you actually give them? :^)

[Anonymous No.15107974]

>>15107969
His solution is as worthless as your skills and thinking process.
Refer to >>15107944
Some minor mistakes may be missed but that's no bad