So, for this problem:
a. Prove that $\sim$ is an equivalence relation on ℝ². (I'm not sure if this is a typo on my professor's part since we are defining a relation on ℕ.)
b. Describe the equivalence class $[1]$.
c. If $p ≠ q$ are primes, then prove that $[p]∩[q] = ∅$.
d. If $n > 1$, then use the fundamental theorem of arithmetic to give a complete description of the equivalence class $[n]$.
So for a, I have that if we let $a\sim a$, we get that $aa = a^2$, and so $a\sim a$ for reflexivity and $\sim$ is reflexive. For symmetry, $ab$ is square, and so is $ba$, so $\sim$ is symmetric. For transitive, let $a\sim b$, and $b\sim c$. Then I get that $ab = x^2$ and $bc = y^2$. So $$ac = \frac{ab\cdot bc}{b^2}=\frac{x^2\cdot y^2}{b^2}=\Big(\frac{xy}{b}\Big)^2 ,$$ and therefore, $ac$ is square, and $\sim$ is transitive.
For b, I get that $[1]$ is by definition the set of all $n \in ℕ$ such that $1\sim n$ or $n$ is square. But that is all I have for it.
For c, can I say that we suppose there is an element, let's say $c$, in the intersection of $[p]$ and $[q]$. Then, $c\sim a$ and $c\sim b$. But since $p$ and $q$ are primes that do not equal each other, then their only possible common element in the intersection is $1$?
For d, I get that the Fundamental Theorem says that every integer $n >1$ can be factored into a product of primes. And so can I say that then $[n]$ is the set of all perfect squares that share the same prime factors of $n$?
So, can anyone tell me if my part a and d are correct, and if anyone can help me with the b, and c?
b) Seems to be correct: you want $1 \times n$ to be a square, so the equivalence class of $1$ is all the prefect squares in $\mathbb{N}$.
c)Hint: assume that there is an element in $[p] \cap [q]$. What can you say about this element, and in particular its divisors?
d) What does the fundamental theorem of arithmetic state? How can you apply this? What is the prime factorisation of a perfect square?