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.

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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.

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.

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.

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.

Sadly we are not allowed complex numbers:

x = i
y = 0
z = -1
works

Originally posted by tomtom232
x=5 y=6 z=13

New to the site but wasn't an almost identical question asked earlier? Just consider the equation mod 4.

Squares of integers are either 0 (if even) or 1 (if odd) mod 4 so the sum of two squares can only be 0, 1 or 2 mod 4 but not 3.