<s>[INST] How to prove this statement. "There are not odd integers $x,y,z$ according to $(x-z)^2+(y-z)^2=(x+y)^2$." I did try to prove this by contradiction(I guess it works). So I  suppose this statement is false but I have confused about the meaning of it.
If this statement is false it means "There are  odd integers $x,y,z$ according to $(x-z)^2+(y-z)^2=(x+y)^2$." or "There are odd integers $x,y,z$ according to $(x-z)^2+(y-z)^2\neq (x+y)^2$." or.....
Is proof by contradiction a good method? and could you help me how to do next from the statement that is supposed to be false?
Thank you very much.
 [/INST] Write $x, y,$ and $z$ as $2a+1, 2b+1,$ and $2c+1$ respectively.
Then, the equation becomes $(2(a-c))^2+(2(b-c))^2=(2(a+b)+2)^2$, or $4(a-c)^2+4(b-c)^2=(2a+2b+2)^2$.
Expanding both sides, one gets $4(a^2-2ac+c^2)+4(b^2-2bc+c^2)=4a^2+4b^2+4+8ab+8a+8b$.
Dividing both sides by $4$, one gets $a^2-2ac+c^2+b^2-2bc+c^2=a^2+b^2+1+2ab+2a+2b$.
The $a^2$ and $b^2$ terms can then be cancelled, giving $-2ac+c^2-2bc+c^2=1+2ab+2a+2b$.
This gives a contradiction, because the LHS is even, while the RHS is odd.
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