No prime p of the form 4k+3 , (p≡3 (mod 4)) is a sum of two squares.

 

\[\begin{array}{l}\\ Modulo\;4,\;\;we\;have\;four\;possible\;remainders\;\\\\ a = 0,1,2,\;\;or\;3\;for\;any\;integer\;a;\;\;consequently,\;\;{a^2} = 0\;or\;1\;(mod4).\\\\ It\;follows\;that,\;\;for\;arbitrary\;integers\;a\;and\;b,{a^2} + {b^2} = 0,\;\;1,or\;2\;(mod4).\\\\ \;Because\;p = 3\;(mod4),\;the\;equation\;p = {a^2} + {b^2}\;is\;impossible.\\ \end{array}\]