Determining the distinct equivalence classes of $3\vert 5a^2+b^2$

Question

Given the equivalence relation R on $\mathbb{Z}$

by $aRb$ iff $3\vert 5a^2+b^2$

Determine the distinct equivalence classes of R

$[0]=\left\{x \in \mathbb{Z}: xR0\right\}=\left\{x \in \mathbb{Z}:3|5x^2+0\right\}$ Would this only give $0$? $[1]=\left\{x \in \mathbb{Z}: xR1\right\}=\left\{x \in \mathbb{Z}:3|5x^2+1\right\}$ this doesn't seem to give me many elements either

Is this the right approach? I figured since this is equivalent to a mod $3$ function it would have $3$ distinct equivalence classes is this not generally right?

What is the definition of $xRy$ here? Is it that $3|5x^2+y^2$? — coffeemath, Dec 05 '16 at 17:22
You actually have three classes, because remainder $2$ can also occur (for example $x=1,y=0$) — Peter, Dec 05 '16 at 17:36

Bill Dubuque · Accepted Answer · 2016-12-05T18:20:45.543

1

Hint $\ {\rm mod}\,\ 3\!:\,\ 0 \equiv b^2\!+5a^2\equiv b^2\!-a^2\equiv (b\!-\!a)(b\!+a\!)\iff \,b\equiv \pm a$

It is not clear what you mean by "since this is equivalent to a mod function it would have $3$ distinct equivalence classes "? Note that equivalence relations on $\{0,1,2\}$ are in bijection with its partitions, so there are some having less than $3$ classes, e.g. the partition $\{0\}\cup \{1,2\}\,$ has $2$ classes.

edited Dec 05 '16 at 18:20

answered Dec 05 '16 at 18:01

Bill Dubuque

272,048

Clever that cleared up a lot for me. Thanks – HighSchool15 Dec 05 '16 at 23:57

score 0 · Answer 2 · answered Dec 05 '16 at 18:03

0

There are two equivalence classes in the set of integers: $A$ is the set of multiples of $3$ and $B$ is the set of integers not divisible by $3.$

It's clear that $xRy$ only depends on $x,y$ mod $3,$ and a check reveals $1R2$ since $5\cdot 1^2+2^2=9.$

answered Dec 05 '16 at 18:03

coffeemath

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Determining the distinct equivalence classes of $3\vert 5a^2+b^2$

2 Answers2