Find all pairs $(x,y)$ of positive integers such that $y^{x^2}=x^{y+2}$.
Problem
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Tags: inequalities, number theory proposed, number theory
pco
22.01.2011 16:43
WakeUp wrote: Find all pairs $(x,y)$ of positive integers such that $y^{x^2}=x^{y+2}$. $x=1\iff y=1$ and the solution $(x,y)=(1,1)$ and so we'll consider from now that $x,y>1$. For any prime $p$, $p|x \iff p|y$. Let then $x_p=v_p(x)$ and $y_p=v_p(y)$ and we get $\frac{x_p}{y_p}=\frac{x^2}{y+2}=\frac uv$ for some coprime natural numbers $u,v$ So $x_p=k_pu$ and $y_p=k_pv$ and $x=a^u$ and $y=a^v$ with $a>1$ and the equation is $va^{2u}=u(a^v+2)$ If prime $p\ne 2|a$, then $v_p(v)+2uv_p(a)=v_p(u)$ and so $2uv_p(a)\le v_p(u)$ and so $2u\le v_p(u)$ which is impossible since $2u>u\ge p^{v_p(u)}\ge v_p(u)$ So $a=2^n$ and the equation is $v2^{2nu-1}=u(2^{nv-1}+1)$ If $nv>1$, we get $2^{nv-1}+1$ odd and so $2^{nv-1}+1|v$ and so $2^{nv-1}+1\le v$ and so $2^{nv-1}\le v-1\le nv-1$ which is impossible ($2^x>x$ $\forall x\ge 1$). So $nv=1$ and $n=v=1$ and the equation is $2^{2u-1}=2u$ and the equation $2^x=x+1$ has only roots $x=0$ and $x=1$ and so $u=1$ So $a=2$ and $u=1$ and $v=1$ and so $x=y=2$ Hence the result $\boxed{(x,y)\in\{(1,1),(2,2)\}}$
vladimir92
27.01.2011 22:48
Case 1. If $x<y$
So $y^{x^2}=x^{y+2}<y^{y+2}$ so $y+2>x^2$. Now rewrite the equation as:
$\left(\frac{y}{x}\right)^x²=x^{y+2-x^2}$ from this and the previous result, follow that $x|y$, write $y=kx^r$ so that $x\nmid k$, the equation becomes then equivalent with: $k^{x^2}=x^{kx^r+2-rx^2}$.
so we get two cases:
=> case 1: if $kx^r+2-rx^2 \geq x^2$ we get that $x|k$ which is impossible.
=> case 2: if $kx^r+2-rx^2<x^2$ then $kx^r+3\leq x^2(r+1)$.
but $kx^r-x^2(r+1)=x^2(kx^{r-2}-(r-2)-3)\geq x^2(k-3+(r-2)(k(x-1)-1))$
so $k=1$ or $k=2$.
If $k=2$, if $x,r\geq 2$, it comes that $2x^r- x^2(r+1) \geq x^2(-1+(r-2)(2x-3))\geq -x^2$ then $x^2\leq 3$ which implies that $x=1$, so $y=2$ but this isn't a solution of the given equation, if checking the cases where $x=1$ or $r=1$, we get again contradictions.
If $k=1$ then $y=x^r$ and thus $x,r\geq 2$ hence: $x^r+2-rx^2=0$
But we have from Bernoulli's inequality:
$x^r-rx^2=x^2(x^{r-2}-r)\geq x^2(1+(x-1)(r-2)-r)=x^2((x-2)(r-2)-1) \geq -x^2$
so $2\geq x^2$ which is impossible.
Case 2. if $x>y$, we can easly get that $x^2\geq y+2$, the proposed equation can be writen as:
$\left(\frac{x}{y}\right)^{y+2}=x^{x^2-y-2}$ so we can write $x=zy^t$ so that $y\nmid z$, and so the proposed equation becomes equivalent with:
$z^{y+2}=y^{z^2y^{2t}-(y+2)(t-2)}$ and like the first part, we have two cases.
=> case 1: $z^2y^{2t}-(y+2)(t-2) \geq y+2$ then $y|z$ which is a contradiction.
=> case 2: $z^2y^{2t}-(y+2)(t-2) \leq y+2$ which is equivalent with: $z^2y^{2t}\leq (y+2)(t-1)$ from Bernoulli's inequality, we have: $y^t \geq 1+(y-1)t=y+(y-1)(t-1)$. But $z^2y^{2t} - (y+2)(t-1)\geq 2y^{t}-1-(y+2)(t-1)\geq 2y-1+(t-1)(y-3)$
so $y\leq 2$.
If $y=2$ it comes that: $4^tz^2\leq 4(t-1)$ which obviously wrong.
If $y=1$ we get $x=1$ which is impossible since $x>y$
Consequently, there is no integers $x\neq y$ so that $y^{x^2}=x^{y+2}$ hence, this equality is satisfied only when $x=y$, meaning that $x^{x^2}=x^{x+2}$ which gives only two solutions: $(x,y)=\{(1,1);(2,2)\}$.
fermat285
28.11.2013 12:55
(1) $x=y=1$ (2) $x,\, y>1$ : There exist positive integer $g>1$ and $a, \, b$ such that $x=g^a,\, y=g^b$, then $bg^{2a}=a(g^b+2)$. If $2a\leq b$, $g^{2a}$ divides $2a$ and it is contradiction, so $2a>b$. Let $2a=b+k$. $bg^{2a}=a(g^b+2)\leq 2ag^b$ so $k+1\geq \frac{b+k}{b}\geq g^{2a-b}\geq 2^{k}$ then $k=1,\, b=1,\, g=2,\, a=1$. So $(x,\, y)=(g^a,\, g^b)=(2,\, 2)$. By (1) and (2), $(x,\, y)=(1,\, 1),\,(2,\, 2)$.