Question

Please explain me this how they did it.

Answer 1

Now first understand an odd number is always 1 greater than some even number

for example
3 is greater than 2
5 is greater than 4

So any odd number can be written as 2q+1
5 = 2(2) +1, 3 = 2(1)+1, 1 = 2(0) + 1

Now above x and y are odd numbers
so
x = 2n+1
y = 2m+1
n,m are integers

now lets solve
$x^{2} + y^2 = (2n+1)^{2}+(2m+1)^{2}$

use 
$(a+b)^2 = a^2+b^2+2ab$
$= (2n+1)^{2}+(2m+1)^{2}$
$= (4n^2 + 4n+1) + (4m^2 + 4m+1)$

Now we have to find out if this is divisble by 4

So any number N = 4q+r, r = 0,1,2,3
for example
10 = 4(2) + 2
12 = 4(3)
14 = 4(3) + 2

As every Number N = 4q + r, r=0,1,2,3
$x^2+y^2 = 4q + r$

[tex](4n^2 + 4n+1) + (4m^2 + 4m+1) = 4q+r \\ 4(n^2+m^2)+4(n+m)+1+1 = 4q+r \\ 4 (n^2+m^2+n+m)+2=4q+r[/tex]

Now n and m are odd so they are not divisible by 4
and same for n^2 and m^2

So sum of squares of any two odd number gives remainder 2 when divided by 4
example
  $1^2+3^2 = 10 = 4(2) + 2$

Hence sum of squares of any two odd numbers is not divisible by 4