2017 - Sum Of Squares?

Fermat's sum of two squares theorem says that any prime that is congruent to 1 mod 4 can be written as the sum of two perfect squares. Since $2016$ is divisible by 4, 2017 satisfies this condition.

It turns out that $44^2 + 9^2 = 2017.$

One of the number is odd and the other is even, say A=2n+1 and B=2m,
(2n+1)^2+4m^2= 2017, 4n^2 + 4n + 1 + 4m^2 = 2017, n(n+1) + m^2 = 504, Since n and n+1 are consecutive numbers, their product is even number and therefore, m is even and m^2 is a multiple of 4, say m= 2q, we have, n(n+1) = 4(126 - q^2), therefore, n(n+1) is of the form, 4p(4p + 1),
So, p (4p + 1) = 126 - q^2, If q is even then p will have to be even number and it can be shown that no integral solution is possible. So q is odd say 2a+1 and then p has to be odd say, 2b+1.
We have, (2b+1)(8b+5) = 126 - (2a+1)^2, 16b^2+18b+5 = 126 - 4a^2 - 4a - 1,
8b^2 + 9b + 2a^2 + 2a = 60,
8b(b+1) + 2a(a+1) + b = 60, a,a+1 and b, b+1, being consecutive numbers, their product is even number and 2a(a+1), 8b(b+1) and 60 being all multiples of 4, the remaining term b has to be a multiple of 4 unless it is zero. For its minimum value b = 4, the LHS becomes more than RHS, so b = 0, then it follows that, 2a(a+1) = 60, Or, a (a + 1) = 30 = 5×6, So a = 5, and we have, A = 2n + 1 = 8p + 1 = 8(2b+1) + 1, = 16b + 9 = 9, and, B = 2m = 4q = 4(2a+1) = 4×11=44, Answer, A = 9 and B = 44.

Well, given from the illustration, because $2017 \equiv1 \mod 4$ then $2017$ can be expressed as a sum of two squares.

it's of the form 4K+1