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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 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.

Lucenaposition
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Poko
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    Hint: You cannot prove uniqueness. – Michael Hardy Feb 25 '14 at 23:37
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    There is a confusion about one point in this question. There is no such object as $\text{“}1\pmod4\text{''}$. When one says that $17$ is $\text{“}$congruent to $1$ $\pmod 4\text{''}$, one is saying that $17$ is congruent mod $4$, to $1$. The object identified by the phrase that includes the words $\text{“mod }4\text{''}$ is the relation of congruence mod $4$. Thus, for example, $17$ and $25$ are congruent-mod-$4$ to each other. ${}\qquad{}$ – Michael Hardy Oct 13 '15 at 15:31
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    $4389=3×7×11×19$ which are all 3 mod 4, but $4389=(3×7)×(11×19)=(3×11)×(7×19)=(3×19)×(11×17)$ so 4389 can be divided into pairs of primes of the form 4n+3 in 3 different ways. – Lucenaposition Sep 10 '23 at 01:54

1 Answers1

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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 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. $

Remark $ $ See here for a generalization to $n$'th powers.

J. W. Tanner
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Bill Dubuque
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