Convert the following integers into sum of two squares:
(a) 202 (b) 205 (c) 180
We will use Sum of Two Squares Theorem (8.5).
Let $m$ be a positive integer given by $m = p_1 \times p_2 \times \ldots × p_r × N^2$ where the $p$’s are distinct primes. Then $m$ can be expressed as a sum of two squares, provided every prime $p_j$ satisfies $p_j = 2$ or $p_j \equiv 1 \pmod 4$ for $j= 1, \ldots , r$.
We will also use the Conversion Identity (8.1).
For any integers $a$, $b$, $c$, and $d$ we have
$$ (a^2 + b^2) (c^2 + d^2) = (ac− bd)^2 + (ad + bc)^2 $$
(a) The prime factorisation is $202=2 \times 101$. We note that $101=1 \pmod 4$.
This gives us
$$ 202 = (1^2+1^2) \times (10^2 + 1^2) $$
Using the Conversion Identity (8.1)
$$ 202 = (10-1)^2 + (1+10)^2 = 9^2 + 11^2$$
(b) The prime factorisation is $5 \times 41$. We note that $5 \equiv 41 \equiv 1 \pmod 4$.
This gives us
$$ 205 = (2^2 + 1^2) \times (5^2 + 4^2) $$
Using the Conversion Identity (8.1)
$$ 202 = (10-4)^2 + (8+5)^2 = 6^2 + 13^2 $$
(c) The prime factorisation of $180 = 2^2 \times 3^2 \times 5 = 6^2 \times 5$. We note that $5 \equiv 1 \pmod 4$.
This gives us
$$ 180 = (6^2 + 0^2) \times (2^2+1^2) $$
Using the Conversion Identity (8.1)
$$ 180 = (12-0)^2 + (6+0)^2 = 12^2 + 6^2$$