- 04 Oct '11 21:17

There are no integers that satisfy.*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.

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. - 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 - 04 Oct '11 22:05

also y^2 = 4z+2 simplifies to (y^2/4) -2 = z*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.