# Thread: Primitive Pyth. triples: Solutions of x^2 + y^2 = 2 z^2

1. Originally Posted by kingwinner
If we want to solve in in the integers, then how would you actually describe all the integer solutions? Does anyone happen to know the correct answer of this problem?

I am solving it in the integers, so the question should say:
Find all INTEGER solutions of $x^2 + y^2 = 2z^2$.

Attempt:
$x^2 + y^2 = 2z^2$
<=> $(x+y)^2 + (x-y)^2 = (2z)^2$

<=>
x+y = +/- $(m^2 - n^2)d$
x-y = +/- $2mnd$
2z = +/- $(m^2 + n^2)d$
where d=0,2,4,6,...,
m and n are any integers of opposite parity with m>n>0 and gcd(m,n)=1
[I say +/- because x+y,x-y,2z may be a Pythagorean triple which requires x+y,x-y,2z to be positive, or some of x+y,x-y,2z may simply be the negatives of it]

<=>
x = +/- $d(m^2 - n^2 + 2mn) / 2$
y = +/- $d(m^2 - n^2 - 2mn) / 2$
z = +/- $d(m^2 + n^2) / 2$
where d=0,2,4,6,...,
m and n are any integers of opposite parity with m>n>0 and gcd(m,n)=1

<=>
x = $d(m^2 - n^2 + 2mn) / 2$
y = $d(m^2 - n^2 - 2mn) / 2$
z = $d(m^2 + n^2) / 2$
where d is any EVEN INTEGER, m and n are any integers of opposite parity with m>n>0 and gcd(m,n)=1

Is this the right way to analyze this problem? Is this the correct description of ALL (up to a reordering of x and y of course) integer solutions of $x^2 + y^2 = 2z^2$?? Did I miss any?
My textbook doesn't give the answer at the back, so there is no way for me to check my work. Can someone confirm that this is correct (or point out any mistake), please?

Thank you very much!
I'd say that's correct.

You could simplify by saying $\frac{d}{2} = a$ then let $a$ be any integer.

2. Originally Posted by kingwinner
hi, I think 9,12,15 is an example of a Pythagorean triple that cannot be written in the form $x = r^2 - s^2, y = 2rs, z = r^2 + s^2$, r,s E Z, r>s>0, right??
Try $a=3,\; r=2,\; s=1$

3. Originally Posted by chiph588@
Try $a=3,\; r=2,\; s=1$
Yes, but (ii) is without the factor a. That's why I'm saying that (i) is right and (ii) is wrong.

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