A strange question about the subset (of $\Bbb N$) of number $4n + 1$; wrong proof? Long introduction (I'm sorry), but very short question 
You take the proper subset of $\mathbb{N}$ made by all and only numbers 
of the form  $4n + 1$, when $n = 0,1,2,\ldots$ and obtain the set :
$A = \{1,5,9,13,17,21,25,29,33,\ldots\}$
$A$ is closed with ordinary product because for any $n$,$m$ of $\mathbb{N}$: 
$(4n+1)(4m+1) = 16nm + 4n + 4m + 1 = 4(4nm+n+m)+1 = 4k + 1 $
Now, in this set $A$, the numbers $9,13,17,21$ (and others) are 
"prime" because they have no factors in $A$. 
$9$  and $21$ are not really prime, of course. They are "prime" in 
the set $A$, because in $A$ there are actually no factors of those
numbers. 
I will call a "prime" in $A$ A-prime (Ap), and for a prime in $\mathbb{N}$ (so, a real prime) simply prime (p). 
Obviously if a number is an Ap not necessarily is a p.
But if it is a p -and belong to $A$- then it is also an Ap.  
Then I will say that a number $Aq$ is a square in $A$ ($A$-square) 
if and only if there exists an $Am$ in $A$ such that $Am^2 = Aq$.
So (for example) the first $A$-square in $A$ is $25$, not $9$.
If $Aq$ is an  $A$-square then  it is also square (in $\mathbb{N}$) but the inverse is false, so the set of the A-square numbers is a proper subset of 
the all squares of $\mathbb{N}$.
Here is the question :
Are there numbers in $A$ not $A$-square (but eventually square), in other words numbers like $Ay$ such that you can find (in $A$) a pair $An^2$, $Am^2$
 that $An^2 = Ay\cdot Am^2$ ?
I refer to the proof kindly offered by Mister Sinclair :
I think is probably (or may be I should say "perhaps") wrong because :
It is true that
$k=3 \pmod 4$ and  $k= 3 \pmod 4$ and it is true that $n = 1 \pmod 4$.
But, although it is NOT true that $1 = 3 \pmod 4$, it IS TRUE that
$1^2=3^2 \pmod 4$ ($1 = 9 \pmod 4$).
Hence, no contradiction rises up.
The problem is (perhaps) a little bit "sneaky" and resembles 
to the following one:
It is TRUE that $(−6)^2 = 3^2\cdot2^2$, so the three number $−6,3,2$ actually 
satisfy the identity.
But it NOT true that $−6=3\cdot2$.
So, I think the problem is not resolved yet
Thank you a lot for any clarification
 A: Pardon me for dropping the $A$ prefixes, which just make the notation clunky. You are looking for integers $n,m,y \equiv 1 \mod 4$ such that $n^2 = y m^2$, but with $y \ne p^2$ for any $p \equiv 1 \mod 4$.
Since these are integers, $y = k^2$ for some $k$. By the condition on $y$, $k \not \equiv 1 \mod 4$. But since $y$ is odd, it must be that $k \equiv 3 \mod 4$. But also, $n = km$, and therefore $n \equiv km \mod 4$. But that means $1 \equiv 3 \cdot 1 \mod 4$, which is false.
Thus no such trio $m, n, y$ can exist. 
A: Let $P_1$ be the set of primes $\equiv 1\pmod 4$ and $P_3$ the set of primes $\equiv 3\pmod 4$. Then we can write any $n\in\mathbb N$ as
$$ n=2^r\cdot\prod_{p\in P_1}p^{s_p}\cdot\prod_{p\in P_3}p^{t_p}.$$
Note that
$$n\in A\iff r=0\text{ and }\sum s_p\text{ is even} $$
$$n\text{ is a perfect square}\iff 2|r\text{, and } 2|s_p\text{ for all $p\in P_1$}\text{, and } 2|t_p \text{ for all $p\in P_3$} $$
$$ n\text{ is an $A$-square}\iff r=0\text{, and all $s_p$ and all $t_p$ are even, and }4\mid \sum s_p$$
Thus the quotient of two $A$-squares (if it is an integer at all) has the following property (obtained by taking the difference of exponents)


*

*$r=0$ because $0-0=0$

*All $s_p$ and all $t_p$  are even because the difference of two evens is even

*$4\mid \sum s_p$ because the difference of two multiples of $4$ is a multiple of $4$.


Hence the quotient of two $A$-squares is an $A$-square or not even an integer.
