This is a question on Pythagorean triples.
Convert the following squares to sum of two non-zero squares:
(a) $5^2$ (b) $17^2$ (c) $29^2$ (d) $202^2$
(a) It is clear that
$$ 5^2 = 3^2 + 4^2 $$
Many people have this Pythagorean triple memorised from school education.
We will use the Conversion Identity (8.1) for the remaining exercises:
$$ (a^2 + b^2) (c^2 + d^2) = (ac− bd)^2 + (ad + bc)^2 $$
(b) We have
$$ 17^2 = (4^2 + 1^2)(4^2 + 1^2) = (16-1)^2 + (4+4)^2 = 15^2 +8^2 $$
(c) We have
$$ 29^2 = (5^2 + 2^2)(5^2 + 2^2) = (25-4)^2 + (10+10)^2 = 21^2 +20^2 $$
(d) We have $202^2=2^2\times 101^2$, and so
$$ 101^2 = (10^2 + 1^2)(10^2 + 1^2) = (100-1)^2 + (10+10)^2 = 99^2 +20^2 $$
And finally,
$$ 202^2 = 2^2 \times (99^2 +20^2 ) = 198^2 + 40^2 $$