<s>[INST] sums of squares and Pythagorean Triples I'm told that if $r^2 + s^2 = z^2$ then $(r+s)^2 + (r-s)^2 = 2z^2$, which is obvious. But i'm trying to show that every integer solution to $x^2 + y^2 = 2z^2$ arises this way from a Pythagorean Triple (r,s,z) and then go on to find the general solution of $x^2 + y^2 = 2z^2$ where $gcd(x,y,z)=1$.
I started by saying that $a,b \in \mathbb{Z}$ was an integer to solution to $x^2 + y^2 = 2z^2$, then $a^2 + b^2 = 2z^2$ and then trying to show that there exist some $a_1 , b_1$ such that $a = a_1 + b_1$, $b=a_1 - b_1$, but i haven't seem to have gotten anywhere. I find myself going in circles. Any tips?
 [/INST] From $a^2+b^2=2z^2,$ assuming $a\ge b,$ if you want $a=a_1+b_1,\ b=a_1-b_1$ then $a_1=(a+b)/2$ and $b_1=(a-b)/2.$ Here you need then to require that $a,b$ have the same parity to be able to do this manipulation.
Note: If one already has $a^2+b^2=2z^2$ then the requirement mentioned above, that the parity of $a,b$ are the same, is implied (an odd square plus an even square would be odd, not $2z^2.$)
An example: $1^2+7^2=2\cdot 5^2,$ and then $(7+1)/2=4,\ (7-1)/2)=3.$ This then goes with $4^2+3^2=5^2.$ So I can see your method will generate Pythagorean triples. SOme care may need to be taken to make sure one ends up with primitive triples.
Added explanation: Starting from $$x^2+y^2=2z^2\tag{*}$$ in which we may assume $\gcd(x,y,z)=1$ and $x \ge y$ for definiteness, we have $x,y$ odd and so may define $r=(x+y)/2,\ s=(x-y)/2.$ Then $$r^2+s^2=(1/2)(x^2+y^2)=(1/2)(2z^2)=z^2.$$
So we do have the solution to $(*)$ arising from a Pythagorean triple $(r,s,z)$ as required.
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