Question

Show that the equation x^2 + y^2 = 4z + 3 has no solution in integers. (Hint: Recall that an even number is of the form 2n and an odd number is of the form 2n+1. Consider all possible cases for x and y even or odd.)

Answer 1

Something like

1) Assume x,y,z are even 2a,2b,2c

(2a)^2 + (2b)^2 = 4(2c) +3

4a^2 + 4b^2 - 8c = 3

4(a^2 +b^2-2c)= 3

and there is no integer you can multiply by 4 to get 3.

Lots of cases....yuk!

Answer 2

Okay! I got that one! :) Thank you once again!! :)

Answer 3

All you have to do now is to solve for the other two situations: both x and y odd, and one even other odd.

Answer 4

Okay~ Thank you!! :)

Answer 5

Ok, you found it already...