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I am reading Ireland's and Rosens's "A Classical Introduction to Modern Number Theory", which contains the following (in my opinion) hard to read argument why every integer of the form $4n + 3$ must be divisible by a prime of the form $4n' + 3$. This is taken from the 2nd edition from page 29:

An integer divided by $4$ leaves a remainder of $0,1,2$ or $3$. Thus odd numbers are either of the form $4k + 1$ or $4k + 3$. The product of two numbers of the form $4k + 1$ is again of that form $$(4k + 1)(4k' + 1) = 4(kk' + k + k') + 1$$ It follows that an integer of the form $4l + 3$ must be divisible by a prime of the form $4l\text{[']} + 3$

Where I have added the ['], since the original text has the form

It follows that an integer of the form $4l + 3$ must be divisible by a prime of the form $4l + 3$

which can be ambiguous. In any case, I do not quite understand why an integer of the form $4n + 3$ has a prime factor of the form $4n' + 3$, and I am a bit confused by the role of the discussion on $4n + 1$ here. I know that sufficiently large primes are all odd and that any odd integer has only odd factors down to the prime factors. Given all of this I do not see the argument for the $4n + 3$ in the given verbatim text.

Bill Dubuque
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Wasradin
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    Look at the contrapositive – if every prime dividing $m$ is of the form $4k+1$, then $m$ must also be of the form $4k+1$, since – the key point – any product of numbers of the form $4k+1$ gives another number of the form $4k+1$. – Gerry Myerson Jan 21 '25 at 11:58
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    @GerryMyerson Thanks, this helped immensely! – Wasradin Jan 21 '25 at 12:53
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    It's just a contrapositive form of integers $\equiv 1$ are closed under multiplication, e.g. see here in the linked dupe, and see the links there for similar arguments. – Bill Dubuque Jan 21 '25 at 20:58

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