75th Romanian Mathematical Olympiad

The equation can be written in the form $(x-y{)}^2+2xy+xy(x-y)=17$. If $x-y=d\in \mathbb{Z}$, $xy=p\in \mathbb{N}$, then the equality becomes $p(d+2)=17-d^2$, so $p=\frac{17-d^2}{d+2}$. We get $p=\frac{13}{d+2}+2-d\in \mathbb{N}$. It follows that $d+2\in \{1,-1,13,-13\}$, which means $d\in \{-1,-3,11,-15\}$. If $d=-1$, then $p=16$ and in this case we have no solutions, because there are no two natural numbers, whose product is $16$ and whose difference is equal to $-1$.

From $d=-3$ we deduce that $p=-8<0$, which doesn't work. For $d=11$, it follows $p=-8<0$, which is false. If $d=x-y=-15$, then $p=xy=16$, whence the unique solution $x=1,y=16$.

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Alternative solution.

The equation becomes successively $(x-y{)}^2+2xy+xy(x-y)=17$ or, if we write it as a difference of two squares: $4(x-y{)}^2+4xy(x-y)+(xy{)}^2-(xy{)}^2+8xy-16=68-16$, that is $(2(x-y)+xy{)}^2-(xy-4{)}^2=52$. It follows that $(2x-2y+xy-xy+4)(2x-2y+xy+xy-4)=52$ and therefore $(x-y+2)(x-y+xy-2)=13$. Since the numbers $x,y,z$ are all natural, there are the following situations: • $x-y+2=-13$ and $x-y+xy-2=-1$, so $x-y=-15$ and $xy=16$, whence $y=x+15$ and $x(x+15)=16$. Since $x$ is a natural number, it follows from this that $x=1$ and then $y=16$; • $x-y+2=-1$ and $x-y+xy-2=-13$, so $x-y=-3$ and $xy=-8$, which is not possible, since $x$ and $y$ are both natural numbers; • $x-y+2=1$ and $x-y+xy-2=13$, so $x-y=-1$ and $xy=16$, which is not possible, since $x$ and $y$ are both natural numbers; • $x-y+2=13$ and $x-y+xy-2=1$, so $x-y=11$ and $xy=-8$, which is not possible, since $x$ and $y$ are both natural numbers.