OK, from 2005 = 5 times 401, obviously two sums of two squares, and
from the multiplicative property

   |(a+bi) (c+di)|^2  =  |a+bi|^2   |c+di|^2,

2005 is the square of the absolute value of
       (2+i)(20+i) = 39 +22 i
and
       (2-i)(20+i) = 41 -18 i

I chose the latter one, as 41 and 18 are sums of two squares too.
What remains is a problem with 3, 4, and 5. Most elegant way should
be to refer to them as the simplest Pythagorean triple, but that
would have been too easy! So I designed the following mess:


      Joyeux Noel                       Merry Christmas

                    et          and

   Meilleurs voeux pour                 Best wishes for

              [y^2+(y+1)^2]^2 + [2x^2]^2   ,

 where  x/y = [sin(pi/x)]^2     ,    y(y+1)=integer part of exp(x)

                                              Alphonse Magnus
                                              magnus@anma.ucl.ac.be
                        see also in http://www.math.ucl.ac.be/~magnus/



Thanks to Laure Ninove and Julien Hendrickx who spotted a mistake in an
early sending.

Andre Ronveaux (and many other people, most probably) performed
wicked reverse engineering on the sums of two squares making 2005,
remarking that there are only two such (integer) sums, from a known
theory (Jacobi).

Bruno Vroman proposed ((x-1)(xy-1))^2+((y-1)(xy+1))^2 for the
other integer sum.

N.Janssens and F.Glineur discussed  real solutions:

try various integers k for the integer part of exp(x): this means
that we have to look for x between log(k) and log(k+1) in the
other equation, where y is replaced by sqrt(k+1/4) -1/2:

  x = [sqrt(k+1/4) -1/2] [sin(pi/x)]^2  ,   log(k) < x < log(k+1),

turning as a standard fixed point problem. For small x, the fixed point
iteration is stabilized through the inverse function:

  x = pi/[a determination of arcsin of +- sqrt(x/y)]
    =  1/[ j  +- Arcsin(sqrt(x/y))/pi]

for integer j. This works for large j, producing pairs close to 1/j,
while k=1, i.e.,  y=(sqrt(5)-1)/2:
 j      x         y
...
10 0.098708251 0.61803399
10 0.10134494 0.61803399
9 0.10943019 0.61803399
9 0.11287389 0.61803399
8 0.12274415 0.61803399
8 0.12738861 0.61803399
7 0.13970938 0.61803399
7 0.14623519 0.61803399
6 0.16204531 0.61803399
6 0.17172478 0.61803399
5 0.19273036 0.61803399
5 0.20820805 0.61803399
4 0.23737394 0.61803399
4 0.26506104 0.61803399
3 0.30775391 0.61803399
3 0.36770246 0.61803399
2 0.43194430 0.61803399

Next solution is x=3/4, y=1  (with k=2).

Then, with j=0, we have five solutions with k=16 to 20:

 k     x         y
16 2.8315364505 3.5311289
17 2.8773674 3.6533119
18 2.9205178 3.7720019
19 2.9613073 3.8874822 
20 3         4

And they discovered completely unsuspected (to me) solutions 

3189 8.0675017 55.973445
3190 8.0679420 55.982298
3191 8.0683822 55.991150

For larger x, y (sin(pi/x))^2 \sim  pi^2 exp(x/2) /x^2 is definitely
larger than x  (exp(x/2) /x^3 is increasing when x>6).

F.Glineur also considered negative values of x, so k=0, and y
must be -1. We now have an infinity of roots of

  x= 1/[j +- Arcsin(sqrt(-x))/pi]

with j a negative integer:
 j      x       y
...
-10 -0.098991474 -1
-10 -0.10104038 -1
-9 -0.10979953 -1
-9 -0.11247152 -1
-8 -0.12324108 -1
-8 -0.12683769 -1
-7 -0.14040494 -1
-7 -0.14544499 -1
-6 -0.16307048 -1
-6 -0.17051744 -1
-5 -0.19435122 -1
-5 -0.20618690 -1
-4 -0.24020886 -1
-4 -0.26114666 -1
-3 -0.31355973 -1
-3 -0.35766104 -1
-2 -0.44775966 -1
-2 -0.57989159 -1
-1 -0.75       -1