This paper is abstracted from 2007 British Mathematics Olympiad Round 1 Question 2.
I am currently practicing grade 8 (Singapore Secondary 2) for the upcoming Singapore Mathematics Olympiad(SMO). Even before solving the question,is it even possible to solve 3 variable of 2 equation?(As from the title of the question).
Question:
Find all solutions in positive integers $x,y,z$ to the simultaneous equations.
$x+y-z=12$
$x^2+y^2-z^2=12$
Of course the first thing that I almost done is $x+y-z=x^2+y^2-z^2$ which is definitely not the way.
I cannot use elimination method as in the end there will be 2 variable in an equation.
I tried using substitution method like $x=12-y+z$,so,$(12-y+z)^2+y^2-z^2=12$ and substituting $x^2=12-y^2+z^2$ where $x=\sqrt {12-y^2+z^2}$,so,$\sqrt {12-y+z}+y-z=12$
Therefore,I get these equations.
- $(12-y+z)^2+y^2-z^2=12$
- $\sqrt {12-y+z}+y-z=12$
If I expand,these equation ended up
- $z^2-yz+y\sqrt {12-y^2+z^2}-z\sqrt {12-y^2+z^2}=78$ (Squared version) OR $\sqrt {12-y^2+z^2}+y-z=12$ (Non-Squared Version)
- $y^2-12y+12z-yz=-66$
From here I don't see how to eliminate one variable away.I am sure what I am doing is wrong or expanded wrongly.Any other ways to do this question?
Best Answer
$$x+y-12=z \Rightarrow x^2+y^2-(x+y-12)^2=12$$
This leads to $$xy-12x-12y+78=0$$ or $$(x-12)(y-12)=66$$
Now check all the possible ways of writing $66$ as a product of 2 integers.