Find all four different representations of $1105= 5 × 13 × 17$ as a sum of two squares.
We will use the Conversion Identity (8.1):
$$ (a^2 + b^2) (c^2 + d^2) = (ac− bd)^2 + (ad + bc)^2 $$
We first note that each prime factor 5, 13, and 17 are congruent to 1 modulo 4, so 1105 can be written as the sum of two squares.
We also note that
$ 5 = 2^2 + 1^2 $
$ 13 = 3^2 + 2^2 $
$ 17 = 4^2 + 1^2 $
Since there are 3 factors, the conversion identity becomes:
$$ \begin{align} (a^2 + b^2) (c^2 + d^2)(e^2 + f^2) & = ((ac− bd)^2 + (ad + bc)^2)(e^2 + f^2) \\ \\ & = (A^2 +B^2)(e^2 + f^2) \\ \\ & = (Ae - Bf)^2 + (Af+Be)^2 \end{align} $$
where $A = (ac− bd)$ and $B = (ad + bc)$.
This could be four different sum of squares by
- swapping $(ac− bd)$ for $(bd-ac)$ which requires swapping $(a^2 + b^2)$ for $(b^2 + a^2)$, and $(c^2 + d^2)$ for $(d^2 + c^2)$.
- swapping $(Ae - Bf)$ for $(Ae - Bf)$ which requires swapping $(A^2 +B^2)$ for $(B^2 +A^2)$, and $(e^2 + f^2)$ for $(f^2 + e^2)$,
The following table shows these calculations.
| product | a | b | c | d | A = (ac − bd) | B = (ad + bc) | e | f | (Ae-Bf) | (Af+Be) | (Ae-Bf)^2+(Af+Be)^2 |
| 5 x 13 | 2 | 1 | 3 | 2 | 4 | 7 | 4 | 1 | 9 | 32 | 1105 |
| 5 x 13 | 1 | 2 | 2 | 3 | 7 | -4 | 1 | 4 | 23 | 24 | 1105 |
| 5 X 17 | 2 | 1 | 4 | 1 | 7 | 6 | 3 | 2 | 9 | 32 | 1105 |
| 5 X 17 | 1 | 2 | 1 | 4 | 6 | -7 | 2 | 3 | 33 | 4 | 1105 |
| 13 x 17 | 3 | 2 | 4 | 1 | 10 | 11 | 2 | 1 | 9 | 32 | 1105 |
| 13 x 17 | 2 | 3 | 1 | 4 | 11 | -10 | 1 | 2 | 31 | 12 | 1105 |
We can read off the four distinct combinations:
$$ \begin{align} 1105 & = 32^2 + 9^2 \\ \\ 1105 & = 24^2 + 23^2 \\ \\ 1105 & = 33^2 + 4^2 \\ \\ 1005 & = 31^2 + 12^2 \end{align} $$