Exercise 3.6.9 (Every prime divisor of $n =a^2+b^2$, where $(a,b) = 1$, satisfies the same property.)

Question

Suppose that $n$ is a positive integer that can be expressed as a sum of two relatively prime squares. Show that every prime divisor of $n$ must also have this property.

Richard Ganaye · Accepted Answer

\begin{proof} Write $$n = 2^\alpha \prod_{p\in A} p^\beta \prod_{q\in B} q^\gamma,$$
 where $A$ is the set of prime divisors of $n$ of the form $4k+1$ and $B$ the set of divisors of $n$ of the form $4k+3$.
 
 If $n$ can be expressed as a sum of two relatively prime squares, then $B = \varnothing$ by Theorem 3.22, so $n = 2^\alpha \prod_{p\in A} p^\beta$. Then every prime divisor of $n$ is $2$ or a prime $p$ of the form $p = 4k+1$. In both cases $p$ is the sum of two relatively prime squares.
\end{proof}

Exercise 3.6.9 (Every prime divisor of $n =a^2+b^2$, where $(a,b) = 1$, satisfies the same property.)

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Exercise 3.6.9 (Every prime divisor of $n =a^2+b^2$, where $(a,b) = 1$, satisfies the same property.)

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