Fermat's eleventh $F_{11}$ represented as sums of two squares? There's an unanswered quest in another forum, that makes me sleepless: Fermat's Elevnth
Let $a,b \in \mathbb{N^{+}}$ and $a>b$ then,
there are $k$-representations for $F_{11}=2^{2^{11}}+1$ as sum of two squares: $ a_{1..k}^{2}+b_{1..k}^{2}$
1.) Find $k$.
2.) Find $a_{1..k}, b_{1..k}$.
Can you help? Thank you!
 A: You need to know the nature of the complete factorization. Evidently $F_{11}$ is odd,squarefree and has exactly five prime factors $p,$ each of which satisfies $p \equiv 1 \pmod 4.$ 
http://www.prothsearch.net/fermat.html#Summary
You do not need to know the specific primes, just the number. So there are 16 expressions as ordered pairs $(x,y)$ of positive numbers, $F_{11} = x^2 + y^2$ 
Oh, the numbers $(x,y)$ are enormous. 
http://www.prothsearch.net/fermat.html

F11 = 319489 · 974849 · 167988556341760475137 · 3560841906445833920513
  · P564

That is enough information to do it all. You can get a computer to specify the prime called $P564$ by getting it to calculate $F_{11},$ carefully typing in the four prime factors given above, then divide them out of $F_{11},$ the final quotient will be the desired $P564.$ As I said in comment, after early progress, this was finished by Richard P. Brent, and $P564$ was proved to be prime by F. Morain, all in 1988. 


A: Hint : If you have $p=a^2+b^2$ and $q=c^2+d^2$, then you have
       $$pq=(a^2+b^2)(c^2+d^2)=(ad-bc)^2+(ac+bd)^2$$
and also $$pq=(a^2+b^2)(c^2+d^2)=(ad+bc)^2+(ac-bd)^2$$
