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I'm struggling something immensely to make sense of the following:

https://meiji163.github.io/post/sum-of-squares/#sums-of-two-squares

Factoring an integer in Gaussian integers is closely related to representing that integer as the sum of two squares. If we can factor $p = (a+bi)(a-bi)$ then $p=a^2 + b^2$. Moreover, if $q = (c+di)(c-di)$ is also the sum of two squares, then so is $pq$ since $$pq = ((ac-bd)+i(ad+bc))((ac-bd)-i(ac+bd)) .$$ From this we can determine all integers that can be represented as a sum of two squares by looking at the primes in its prime factorization (in regular integers). For example, $15=3 \cdot 5$ is not the sum of squares because we can’t factor the $3$ in Gaussian integers.

This should be easy, but for some reason my mind is still stuck somewhere on some trivial point. I just don't understand how you get from "if $x$ and $y$ both have property A, then so does their product $xy$" to "if $z$ has property $A$ and $x$ is a prime factor of $z$, then $x$ also has property $A$".

Someone please save me from this fog of ignorance!

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    $\begingroup$ The thing that addresses separate factors: if $x^2 + y^2 \equiv 0 \pmod q $ for (positive) prime $q \equiv 3 \pmod 4,$ then both $x,y \equiv 0 \pmod q,$ so that $x^2 + y^2 \equiv 0 \pmod {q^2} $ Induction says that the exponent for $q$ must be even. $\endgroup$
    – Will Jagy
    Commented Feb 16 at 16:55
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    $\begingroup$ It's not true that if $z$ is the sum of two squares and $x$ is a prime factor of $z$, then $x$ is the sum of two squares. For example, $9 = 0 + 3^2$ is the sum of two squares, but $3$ is not the sum of two squares. $\endgroup$
    – Robert Israel
    Commented Feb 16 at 16:56
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    $\begingroup$ This blog does seem to have skipped some of the usual justification given! It's not a totally trivial fact indeed, and in this case it may be that $z$ can be written as a sum of squares but a prime factor cannot (eg $z = 45$, $p = 3$). The argument is a bit more subtle than that - you can find it eg as Theorem 6 here or in section 2.6 here. $\endgroup$
    – Izaak van Dongen
    Commented Feb 16 at 17:00
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    $\begingroup$ There is a middle ground when the (form) class number is bigger than one. For discriminant $-20$ we get something similar for $x^2 + 5 y^2.$ This obviously represents $6$ with integers $x=1, y=1.$ Of course, $x^2 + 5 y^2$ does not represent $2$ or $3.$ However, the other form $2 x^2 + 2 x y + 3 y^2$ obviously does represent both $2,3$ In general, if $m,n$ are both represented by $2 x^2 + 2 x y + 3 y^2,$ there is also a representation of the product, $x^2 + 5 y^2 = mn.$ Along with $2,$ $2 x^2 + 2 x y + 3 y^2$ represents all primes $p \equiv 3,7 \pmod{20}$ $\endgroup$
    – Will Jagy
    Commented Feb 16 at 17:20

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