TITLE: When are $n$ and $2n$ both sums of two squares?
QUESTION [9 upvotes]: Given a natural number $n$, can we completely characterize when $n$ and $2n$ are each a sum of two squares?
For example: $446,382,709=(13010)^{2}+(16647)^{2}$ and $892,765,418=2(446,382,709)=(3637)^{2}+(29657)^{2}$
Do these imply, perhaps, that 446,782,709 is the hypotenuse of a primitive pythagorean triple?
(I found this question on a slip of paper while cleaning my office. It turns out that a trivial algebraic identity resolves the question...beyond Euler's characterization of when a natural number is the sum of two squares. Is the question nontrivial if we replace $2n$ by $3n$ above?)
EDIT: The modification is trivial, too. This is my fault, as I didn't think about my question before posting it. (This was a problem found on a slip of paper in my office. I posted it because it looked like a fun problem for students. This is certainly not the best forum for such things!!!)

REPLY [19 votes]: A natural is the sum of two squares iff every prime of the form $\rm\:4\:k+3$ occurs to even power in its prime factorization (iff the same is true for $\rm\:2\:n\:$). Note $\rm\ n = x^2 + y^2\ \Rightarrow\ 2\:n = (x+y)^2 + (x-y)^2$ which arises by compositions of forms, or, in linear form, with the norm $\rm\ N(a+b\ i)\ =\ a^2 + b^2 $
$$\rm 2\ (x^2+y^2)\ =\ N(1+i)\ N(x-y\ i)\ =\ N((1+i)\ (x-y\ i))\ =\ N(x+y + (x-y)\ i) $$

REPLY [14 votes]: $2(a^2 + b^2) = (a + b)^2 + (a - b)^2$.  And, of course, the classification of numbers which are the sums of two squares is well-known (it is necessary and sufficient that if $p \equiv 3 \mod 4$, then the largest $k$ such that $p^k | n$ is even).