Estonian Mathematical Olympiad

Substituting $z=x+y-23$ from the first equation into the second yields $x^2+y^2-(x+y-23{)}^2=23$ which simplifies to $xy-23x-23y=-23\cdot 12.$ Adding $23\cdot 23$ to both sides and factoring yields $$(x-23)(y-23)=23\cdot 11.$$ As $23$ and $11$ are primes, the only factors on the right hand side are $1$, $11$, $23$ and $11\cdot 23$. Thus $x=23+1=24$, $x=23+11=34$, $x=23+23=46$ or $x=12\cdot 23=276$; The corresponding values of $y$ are $276$, $46$, $34$, $24$ and the values of $z$ are $277$, $57$, $57$, $277$. The negative factors of $23\cdot 11$ don't yield solutions, as $z$ would be negative.

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Alternative solution.

Substituting $z=x+y-23$ from the first equation into the second and simplifying like in Solution 1 yields $$xy=23x+23y-23\cdot 12.\text{\hspace{1em}(1)}$$ The right hand side of (1) is divisible by $23$, so the left side must be as well. Thus $x$ or $y$ is divisible by $23$. Let $x=23k$ (in the other case we may swap $x$ and $y$ due to symmetry). Substituting it into (1) and dividing the sides by $23$ yields $ky=23k+y-12$, from which $k=\frac{y-12}{y-23}=1+\frac{11}{y-23}$. For $k$ to be a positive integer, there are two possibilities: if $y-23=11$, then $k-1=1$, which yields $y=34$, $x=46$, $z=57$; if $y-23=1$, then $k-1=11$, which yields $y=24$, $x=276$, $z=277$. In addition to those, we get the solutions with $x$ and $y$ swapped.