Originally posted by talzamir Are there integers for which x^2 + y^2 = 4z + 3?
If yes, give an example of such a triplet of numbers.
If no, explain why such a triplet doesn't exist.
There are no integers that satisfy.
Because 1 is the only integer that multiplied by itself yield an odd number, and 4z + 3 must be an odd number, either x or y must be 1.
So, because x and y are interchangeable, let's assume that x is 1.
this yields:
1 + y^2 = 4z + 3
y^2 = 4z +2
this yields
(y^2 -2)/4 = z
this yields
z = (1/4)y^2 - 1/2
Any y integer will yield a remainder of 1/2 , so therefore, z cannot be an integer.
Originally posted by TomCr There are no integers that satisfy.
Because 1 is the only integer that multiplied by itself yield an odd number, and 4z + 3 must be an odd number, either x or y must be 1.
So, because x and y are interchangeable, let's assume that x is 1.
this yields:
1 + y^2 = 4z + 3
y^2 = 4z +2
this yields
(y^2 -2)/4 = z
this yields
z = (1/4)y^2 - 1/2
Any y integer will yield a remainder of 1/2 , so therefore, z cannot be an integer.
Because 1 is the only integer that multiplied by itself yield an odd number
Originally posted by TomCr There are no integers that satisfy.
Because 1 is the only integer that multiplied by itself yield an odd number, and 4z + 3 must be an odd number, either x or y must be 1.
So, because x and y are interchangeable, let's assume that x is 1.
this yields:
1 + y^2 = 4z + 3
y^2 = 4z +2
this yields
(y^2 -2)/4 = z
this yields
z = (1/4)y^2 - 1/2
Any y integer will yield a remainder of 1/2 , so therefore, z cannot be an integer.