I was surfing the web when I stumbled upon this given problem. I tried to solve this problem using high school math, and I just couldn't cut it.
My question is how do you solve this problem? If you could answer, is there any techniques or strategies to be solving problems similar to this one?
Thank you very much.
$$-1=a^2+b^2-(a+b-1)^2=-2ab-1+2a+2b,$$ which gives $$ab=a+b$$ or $$(a-1)(b-1)=1.$$ Can you end it now?
I got $$\{(0,0,-1),(2,2,3)\}$$ For example, $$(a-1)(b-1)=1$$ gives $$a-1=1$$ and $$b-1=1$$ or $$a-1=-1$$ and $$b-1=-1.$$ In the first case we obtain $a=b=2$ and since $a+b-c=1,$ we obtain: $$2+2-c=1$$ or $$c=3.$$