rst
Youngling
Re: mathematic related questions
But you can use this method only with number giving remainder 3 (when divided by 4)
If a number gives remainder 0,1,2 ,then x² + y²=number may have solution or not
Also with this method we can't find number of solutions of x² + y²=number (in case it has solution)
see my first post (page 1)
BTW if I had to solve such questions i would have probably applied hit and trial using the following conditions :
But you can use this method only with number giving remainder 3 (when divided by 4)
If a number gives remainder 0,1,2 ,then x² + y²=number may have solution or not
Also with this method we can't find number of solutions of x² + y²=number (in case it has solution)
offtopic : do you guys know what is the shortcut key for subscript and superscript here in this forum ?
have to go advanced everytime
see my first post (page 1)