Find all $ x,y$ and $ z$ in positive integer: $ z + y^{2} + x^{3} = xyz$ and $ x = \gcd(y,z)$.
Problem
Source: IMO Shortlist 1995, N4
Tags: algebra, number theory, equation, greatest common divisor, IMO Shortlist
Myth
07.01.2005 00:02
Also Russian TST 1997 (where I was qualified for IMO)
Myth
20.10.2006 12:00
Anyway, some one should write it. Note, then ehsan2004 made a typo. The right condition is $x=(y,z)$. So, let $y=xy'$ and $z=xz'$. Then $z'+xy'^{2}+x^{2}=x^{2}y'z'$. It means $x\mid z'$, hence $z'=xz''$ (also $z=x^{2}z''$). It follows that $z''+y'^{2}+x=x^{2}yz''$. We have obtained quadratic equation with respect to $y'$. It has a quadratic discriminant $D^{2}=x^{4}z''^{2}-4(x+z'')$, where $D\in Z$. Therefore, \[4(x+z'') \geq 4x^{2}z''-4.\]But $x^{2}z''-(x+z'')-1=(xz''-z''-1)(x+1)$. Thus $xz''-z''-1\leq 0$. It means that $z''=1$ OR $x=1$. If $z''=1$ then $D^{2}=x^{4}-4x-4=(x^{2}-2)^{2}+(4x^{2}-4x-8)$. So, for $x>2$ we obtain $x^{2}-2<D<x^{2}$ and $D$ has the same parity as $x$. Contradiction. And for $x=1$ we obtain $D^{2}=-7$ and for $x=2$ we obtain $D^{2}=4$, hence $y'\in\{1,3\}$. If $x=1$ then $D^{2}=z''^{2}-4z''-4=(z''-2)^{2}-8$. Hence $z''-2=3$. Finally, the answer is $x=1$, $y\in\{2,3\}$, $z=5$ OR $x=2, y=2, z=4$ OR $x=2,y=6,z=4$. P.S. I have corrected a mistake where I forget a case $x=2$.
FelixD
08.04.2009 14:41
What about $ (2,2,4)$ and $ (2,6,4)$?
Jorge Miranda
05.01.2010 13:46
First of all note that $ x\mid y$ and $ x^2\mid z$ (because $ x\mid y\Rightarrow x^2\mid xyz$ e $ x\mid y\Rightarrow x^2\mid y^2$). So let $ x=a$, $ y=bx$, $ z=cx^2$, with $ a\perp b\perp c$
The given equation is then equivalent to $ a^3+(ab)^2+(a^2c)=a^4bc\Leftrightarrow a+b^2+c=a^2bc\Rightarrow a^2b-1\mid a+b^2$.
But then we have that $ a^2b-1\mid a^3+b=(a+b^2)a^2-b(a^2b-1)$
From that we have that $ a^2b-1\leq a^3+b\Leftrightarrow a=1 \vee b\leq a+\frac{1}{a-1}$
If $ a=1$, $ a^2b-1\mid a^3+b\Leftrightarrow b-1\mid b+1\Leftrightarrow b-1\mid 2\Leftrightarrow b-1\in\{1,2\}\Leftrightarrow b\in\{2,3\}$
From this we have $ c=\frac{a+b^2}{a^2b-1}=5$ in both cases, so $ (x,y,z)\in\{(1,2,5),(1,3,5)\}$, which are solutions.
If $ a=2$, then $ b\leq 3$ and $ a^2b-1\mid a^3+b \Leftrightarrow 4b-1\mid b+8$, from where we conclude that $ b\in \{1,3\}$, so $ c=\frac{a+b^2}{a^2b-1}=1$ in both cases. From this we have $ (x,y,z)\in\{(2,2,4),(2,6,4)\}$, which are solutions.
If $ a>2$, $ b\leq a+\frac{1}{a-1}\Rightarrow b\leq a$ (because $ b\in\mathbb{N}$). We have that $ a^2b-1\mid a+b^2\Rightarrow a^2b-1\leq a+b^2\Leftrightarrow -ba^2+a+b^2+1\geq 0$. But $ -ba^2+a+b^2+1$ is decreasing w.r.t $ a$ (it's derivative w.r.t $ a$ is $ -2ab+1<0$ for example), so if $ a\geq b+1$ we have that $ -ba^2+a+b^2+1\leq 2-b^2-b^3<0$, contradiction (unless $ b=1$ e $ a=b+1=2$, contradiction again). Otherwise we have $ a=b$, but then $ a\not\perp b$, contradiction.
We conclude that the only solutions are $ (x,y,z)=(1,2,5),(1,3,5),(2,2,4),(2,6,4)$
RishiNandha_M
26.02.2020 19:12
I did Something but somehow missed the even cases. (I did the same ques with x and z letters alone switched. $x = hz; y = kz$ h,k are coprimes. $h + zk^2 = z^2(hk - 1)$ $z | h+zk^2 \ ; \ z| \frac hz + k^2 $ $z | h \ ; \ h=gz$ Then, $z | g + k^2$ Since h and k were coprimes, $z !| k$ and $z !| k^2$ Therefore $g = fz$ Original Equation becomes $$z(fkz^2-f-1) = k^2$$h and k were coprimes and $z|h$, Therefore $z !| k$ Only possibility if $$z=1 \ ; \ fk - f - k^2 - 1 = 0$$, Taking quadratic in k, $k = \frac {f \pm \sqrt{(f-2)^2-8}}2$ $(f-2)^2-8 = a^2$ $$8 = (f-2-a)(f-2+a) ; (f,k,z) = (f, \frac {f \pm a}2,1)$$$(a+f-2,f-2-a) = [(2,4),(4,2),(-2,-4),(-4,-2)]$, we take only same parity duples. $(f,a)=[(5,-1),(5,1),(-1,1),(-1,-1)]$ And $f=a \mod 2$ for k to be an integer and $f>0$ $(f,k)=[(5,1),(5,-1)]$ and k>0 given. And also f and k have to be coprimes as h and k are coprimes and z=1, So $$(f,k,z) = [(5,3,1),(5,2,1)]$$ $$(x,y,z) = [(5,3,1),(5,2,1)]$$are the $Z^+$ solutions to $x + y^2 + z^3 = xyz$ where $z = gcd(z,y)$. Help me Where I missed out on Considering Other two cases PLEAASEEE