Let $H=\{1,5,9,\cdots\}$ be the arithmetic progression of all positive integers of the form $4k+1$.Elements of $H$ are called Hilbert numbers.Show that $H$ is closed under multiplication,that is ,$x,y\in H$ implies $xy\in H$.An element $x$ of $H$ will be called Hilbert prime if $x\neq 1$ and $x$ can not be written as the product of two strictly smaller elements of $H$.Compute all the Hilbert primes up to 100.Prove that every element of $H$ can be factored into a product of Hilbert primes,but that unique factorization does not hold in $H$.
Proof:
(1)$H$ is closed under multiplication:
\begin{equation}
(4k_1+1)(4k_2+1)=16k_1k_2+4(k_1+k_2)+1=4[4k_1k_2+k_1+k_2]+1
\end{equation}Done.
(2)5,9,13,17,21,29,33,37,41,49,53,57,61,69,73,77,89,93,97,101,109,113,121,129,……
(3)The existence of the factorization is simple.The non-uniqueness,please see here.