#### Posers and Puzzles

1.  talzamir
Art, not a Toil
28 Sep '11 16:00
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.
2. 28 Sep '11 16:18
No, consider the equation mod 4
3.  TomCr
woodpusher
04 Oct '11 21:17
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.
4. 04 Oct '11 21:38 / 1 edit
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

3*3=9, 5*5=25 etc etc ad infinitum
5. 04 Oct '11 21:42
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.
x=5 y=6 z=13
6. 04 Oct '11 22:05
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.
also y^2 = 4z+2 simplifies to (y^2/4) -2 = z
7. 04 Oct '11 23:27
Sadly we are not allowed complex numbers:

x = i
y = 0
z = -1
works
8.  TomCr
woodpusher
05 Oct '11 00:54
Originally posted by tomtom232
x=5 y=6 z=13
5x5 + 6x6 = 4x13 + 3
25 + 36 = 52 + 3
61 is not 55.

5,6,13 does not work.
9. 05 Oct '11 05:23
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.