Convert the following integers into sum of two squares:
(a) 2016
(b) 2015
(c) 2017
Hint: 2017 is prime. Consider the square root of 2017.
(d) 2018
Hint: $2018= 2 \times 1009$ and determine $1009− 28^2$.
(e) $2019= 3 \times 673$
(f) 2020
(a) The prime factorisation is $2016 = 2^5 × 3^2 × 7$.
Since $7 \equiv 3 \pmod 4$ and appears with an odd power, by Corollary (8.8) the number 2016 can't be written as a sum of two squares.
(b) The prime factorisation if $2015 = 5 \times 13 \times 31$.
Since $31 \equiv 3 \pmod 4$ and appears with an odd power, by Corollary (8.8) the number 2015 can't be written as a sum of two squares.
(c) Taking the hint $\lfloor \sqrt{2017} \rfloor = 44$ gives us
$$ 2017 = 44^2 + 9^2 $$
(d) The prime factorisation is $2018= 2 \times 1009$. Taking the hint we have $1009-28^2=15^2$. And so
$$ 2018 = (1^2+1^2) \times (28^2 + 15^2) $$
By the Conversion Identity (8.1)
$$ 2018 = (28-15)^2 +(15+28)^2 = 13^2+43^2$$
(e) The prime factorisation is $2019=3 \times 673$.
Since $3 \equiv 3 \pmod 4$ and appears with an odd power, by Corollary (8.8) the number 2019 can't be written as a sum of two squares.
(f) The prime factorisation is $2020=2^2 \times 5 \times 101$.
Rewriting,
$$ 2020 = 2^2 \times (2^2 + 1^2) \times (10^2 + 1^2) = (4^2 + 2^2) \times (10^2 + 1^2) $$
By the Conversion Identity (8.1)
$$ 202 = (40-2)^2 + (4+20)^2 = 38^2 + 24^2 $$