Which unfactored sofar RSA numbers have sum of squares representation?

[hermann@stamm-wilbrandt.de]

A requirement for sum of squares representation of n=RSA-l is, that 
n%4==1.

Here are all (un)factored sofar RSA numbers that are =1 (mod 4).
Separation possible by nrp = Mod(qnr, n)^(phi/4), with qnr a quadratic 
non-residue (mod n), and phi=(p-1)*(q-1).
But knowing n=p*q is required.
If p==1 (mod 4) and q==1 (mod 4), p and q have unique sum of squares.
And by Brahmagupta–Fibonacci identity n has two representations as sum 
of squares.
If p==3 (mod 4) and q==3 (mod 4), n does not have sum of squares 
representation.

pi@pi400-64:~/RSA_numbers_factored/pari $ gp -q < phi.gp
validate(rsa): ✓
RSA p%4 q%4 nrp
 59   1   1   1
129   1   1   1
130   3   3  -1
155   3   3  -1
160   3   3  -1
180   1   1   1
190   3   3  -1
640   3   3  -1
220   3   3  -1
230   1   1   1
768   1   1   1
250   3   3  -1
280
309
310
320
340
360
370
390
430
450
460
1536
470
500
617
2048
pi@pi400-64:~/RSA_numbers_factored/pari $



pi@pi400-64:~/RSA_numbers_factored/pari $ cat phi.gp
\r RSA_numbers_factored
print("RSA p%4 q%4 nrp");
for(i=1,#rsa,\
  if(#rsa[i]>=4,\
    [l,n,p,q]=rsa[i];\
    forprime(t=2,oo,if( kronecker(t, p)==-1, qnrp=t; break()));\
    assert(Mod(qnrp, p)^((p-1)/2) == Mod(-1, p));\
    forprime(t=2,oo,if( kronecker(t, q)==-1, qnrq=t; break()));\
    assert(Mod(qnrq, q)^((q-1)/2) == Mod(-1, q));\
    m = chinese(Mod(qnrp, p), Mod(qnrq, q));\
    qnr = lift(m);\
    assert(Mod(qnr, n) == m);\
    phi=(p-1)*(q-1);\
    assert(Mod(qnr, n)^(phi/2) == 1);\
    if(n%4==3, next);\
    printf("%3d %3d %3d %3d\n", l, p%4, q%4, centerlift(Mod(qnr, 
n)^(phi/4))),\
    [l,n]=rsa[i];if(n%4==1,print(l));\
  )\
)
pi@pi400-64:~/RSA_numbers_factored/pari $



Regards,

Hermann.