<s>[INST] Solving the simultaneous equations $x + y - z = 12$ and $x^2 + y^2 - z^2 = 12$ I found the following question online (it was a past question on the BMO):
find all positive integer solutions $x,y,z$ solving the simultaneous equations 
$ x + y - z = 12 \ $ and $\ x^2 + y^2 - z^2 = 12 $   .
Normally these questions have a simple and elegant solution and although I've managed to solve it, the answer I found was neither! I wondered if anyone can see a simpler way to tackle the problem.
The way I solved this was to use the first equation to eliminate $z$ in the second question. I then factorised the resulting equation to obtain:
$(x-12)(y-12) = 66$ .
Then, using the fact that $66=2\times 3\times 11$ and also the fact that the equations are symmetric in $x$ and $y$, I worked out the possible factorisations of $66$ into two factors. From this I worked out possible solutions and then checked to see that they indeed were. (In total I found $8$ solutions.) Although this method worked, it seemed a bit cumbersome and I hope someone can improve on it! Please note that this question is aimed at someone with at most A-level standard maths abilities meaning it shouldn't need complex mathematics to solve it!
 [/INST] I'll try a generalization.
From
$x+y-z = n
$
and
$x^2+y^2-z^2 = m
$,
$z = x+y-n$
so
$z^2
=(x+y)^2-2n(x+y)+n^2
=x^2+2xy+y^2-2n(x+y)+n^2
$.
Therefore
$m
=x^2+y^2-(x^2+2xy+y^2-2n(x+y)+n^2)
=-2xy+2n(x+y)-n^2
$
so
$xy-n(x+y)
=-(m+n^2)/2
$
or
$(n^2-m)/2
=xy-n(x+y)+n^2
=(x-n)(y-n)
$.
The integer solutions
thus depend on the
factors of
$N=(n^2-m)/2$.
Note that if
$n$ and $m$ have different parity,
there are no integer solutions.
For each
$N
=ab
$
either
$x=n+a,
y=n+b
$
or
$x=n-a,
y=n-b
$.
The largest solution
is gotten by setting
$a=N$ and $b = 1$
so
$x = n+N$
and
$y = n+1$.
</s>