Modular arithmetic question related to the fundamental theorem Somewhat of an unusual homework problem that my professor assigned that I can't wrap my head around.
We are only considering the positive numbers congruent to 1(mod 4), that is, other numbers do not exist in this problem. A number is said to be "primey" if its only positive divisors congruent to 1(mod 4) are 1(mod 4) and itself (For example: 9 is primey however 9 is not prime). The question is: is it true that every number congruent to 1(mod 4) different than 1 is the product of "primey" numbers uniquely?
I have shown every number different than 1 and congruent to 1(mod 4) can be expressed as a product of primeys. In fact, it was very similar to the proof showing the same for the natural numbers. However, I am having trouble with the uniqueness part. I know that the uniqueness does not hold here, but I do not know why. 
Thanks in advance. 
 A: Hint $ $ Suppose we have primeys (irreducibles) $\,a^2\ne b^2$ where $\,a,b\equiv -1\pmod{\! 4}.\,$ Multiplying yields $\,a^2 b^2 = (ab)^2,\, $ so their product is a square in this number system, since $\,ab\equiv 1\pmod{\!4}.\,$ But the product of two nonassociate irreducibles cannot be a square if factorization is unique.
Remark $ $ This is a famous example Hilbert used to motivate the restoration of unique factorization using ideals (or "ideal factors"). By the Lemma below, if we have unique factorization then coprime factors $A,B\,$ of a square must themselves be squares $\ AB = C^2$ $\Rightarrow$ $A = (A,C)^2.$ Were factorization unique, we'd deduce from $\,9\cdot49 = 21^2\, $ that $\,9 = (9,21)^2\, $ and $\,49 = (49,21)^2.\,$ Adjoining these gcds as new ("ideal") elements eliminates this nonunique factorization since then the two factorizations have the common refinement $\,(9,21)^2\, (49,21)^2 = \color{#90f}{3^2 7^2}.\,$ In effect we have discovered the new ("$\rm\color{#90f}{ideal}$") elements $\, \color{#90f}3= (9,21)\,$ and $\,\color{#90f}7 = (49,21)\,$ needed to restore unique factorization by taking gcds of ("real") elements in our number system $\,1+4\,\Bbb N.\,$
Lemma $\rm\ \ \color{#0a0}{(a,b,c) = 1},\,\ \color{#c00}{c^2 = ab}\ \Rightarrow\ a = (a,c)^2,\,\ b = (b,c)^2\ $ for $\rm\:a,b,c\in \mathbb N$
Proof $\rm\ \ (c,b)^2 = (\color{#c00}{c^2},b^2,bc) = (\color{#c00}{ab},b^2,bc) = b\color{#0a0}{(a,b,c)} = b.\ $ Similarly for $\,\rm(c,a)^2. $
A: Just a hint:
When I was an undergraduate a friend of mine saw this same problem in a number theory course he took, except that it was phrased differently: Find the smallest number congruent to $1\bmod 4$ whose factorization of this kind is not unique.
