求正整数数对 (x,y)(x,y)(x,y) ,满足 x>yx>yx>y 且
(x−y)xy=xyyx. (x-y)^{xy}=x^yy^x.(x−y)xy=xyyx.
(2013,中国台湾数学奥林匹克训练营)
解法1:
设
(x,y)=d,x=dv,y=dt,(v,t)=1.(x,y)=d ,x=dv,y=dt,(v,t)=1.(x,y)=d,x=dv,y=dt,(v,t)=1.
则:
dxy−x−y(v−t)=vytxd^{xy-x-y}(v-t)=v^yt^xdxy−x−y(v−t)=vytx
⇒d(x−1)(y−1)−1(v−t)xy=vytx\Rightarrow d^{(x-1)(y-1)-1}(v-t)^{xy}=v^yt^x⇒d(x−1)(y−1)−1(v−t)xy=vytx
若 (x−1)(y−1)−1=−1(x-1)(y-1)-1=-1(x−1)(y−1)−1=−1
⇒x=1(舍去)或y=1\Rightarrow x=1(舍去)或y=1⇒x=1(舍去)或y=1
⇒(x−1)x=x\Rightarrow (x-1)^x=x⇒(x−1)x=x
⇒无正整数解\Rightarrow 无正整数解⇒无正整数解
若 (x−1)(y−1)−1≥0(x-1)(y-1)-1\geq0(x−1)(y−1)−1≥0 ,则由
(v−t,v)=1,(v−t,t)=1(v-t,v)=1,(v-t,t)=1(v−t,v)=1,(v−t,t)=1,
而 (v−t)xy∣vytx(v-t)^{xy}|v^yt^x(v−t)xy∣vytx , 知
v−t=1v-t=1v−t=1
故 dd(t+1)t−(2t+1)=(t+1)ttt+1d^{d(t+1)t-(2t+1)}=(t+1)^tt^{t+1}dd(t+1)t−(2t+1)=(t+1)ttt+1.
若 ttt 由素因子,则对 ttt 的任一素因子 ppp ,设
pα∣∣t,pβ∣∣dp^{\alpha}||t,p^{\beta}||dpα∣∣t,pβ∣∣d (α、β≥1).(\alpha、\beta\geq1).(α、β≥1).
则 β[d(t+1)t−2t−1]=α(t+1)⋯①\beta[d(t+1)t-2t-1]=\alpha (t+1) \cdots ①β[d(t+1)t−2t−1]=α(t+1)⋯①
⇒(βdt−α)(t+1)=(2t+1)β.\Rightarrow (\beta dt-\alpha)(t+1)=(2t+1)\beta.⇒(βdt−α)(t+1)=(2t+1)β.
而(t+1,2t+1)=1,(t+1,2t+1)=1,(t+1,2t+1)=1, 故 (t+1)∣β.(t+1)|\beta.(t+1)∣β.
由式①知
β∣α(t+1)\beta |\alpha(t+1)β∣α(t+1)
⇒β≤α(t+1)\Rightarrow \beta \leq \alpha(t+1)⇒β≤α(t+1)
⇒(βdt−α)(t+1)≤(2t+1)α(t+1)\Rightarrow (\beta dt-\alpha)(t+1)\leq(2t+1)\alpha(t+1)⇒(βdt−α)(t+1)≤(2t+1)α(t+1)
⇒βdt≤2(t+1)α≤2βα\Rightarrow \beta dt \leq2(t+1)\alpha\leq2\beta\alpha⇒βdt≤2(t+1)α≤2βα
⇒2pα≤dt≤2α⇒pα≤α\Rightarrow 2p^{\alpha}\leq dt\leq2\alpha\Rightarrow p^{\alpha}\leq\alpha⇒2pα≤dt≤2α⇒pα≤α
⇒无正整数解\Rightarrow 无正整数解⇒无正整数解.
矛盾.
因此t=1⇒d2d−3=2⇒d=2t=1\Rightarrow d^{2d-3}=2\Rightarrow d=2t=1⇒d2d−3=2⇒d=2
故(x,y)=(4,2).(x,y)=(4,2).(x,y)=(4,2).
解法2:
若 p∣xp|xp∣x , ppp 为素数,则
p∣(x−y)⇒p∣y.p|(x-y) \Rightarrow p|y.p∣(x−y)⇒p∣y.
因为 x≥2x\geq2x≥2 ,所以,这样 ppp 不存在.
故y≥2y\geq 2y≥2.
类似地,若 p∣yp|yp∣y ,则 p∣xp|xp∣x .
不妨设 x=p1α1p2α2⋯pkαkx=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}x=p1α1p2α2⋯pkαk ,y=p1β1p2β2⋯pkβky=p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}y=p1β1p2β2⋯pkβk .
若存在正整数 iii ,使得αi<βi\alpha_i<\beta_iαi<βi ,则考虑 pip_ipi 在方程两边的幂次,得
αixy=αiy+βix<βi(x+y)\alpha_ixy = \alpha_iy+\beta _ix<\beta_i(x+y)αixy=αiy+βix<βi(x+y)
<y(x+y)<2xy<y(x+y)<2xy<y(x+y)<2xy
⇒αi=1.\Rightarrow \alpha_i=1.⇒αi=1.
故 xy=y+βix<x+βixxy=y+\beta_ix<x+\beta_ixxy=y+βix<x+βix
⇒βi+1>y≥piβi≥2βi\Rightarrow \beta_i+1>y\geq p_i^{\beta_i}\geq2^{\beta_i}⇒βi+1>y≥piβi≥2βi
⇒\Rightarrow⇒ 无解.
于是,对于任意整数 i(1≤i≤k)i(1\leq i\leq k)i(1≤i≤k) , 均有
αi≥βi⇒y∣x\alpha_i\geq\beta_i\Rightarrow y|xαi≥βi⇒y∣x.
设 x=tyx=tyx=ty (t≥2)(t\geq2)(t≥2).则
(t−1)ty2yty2=tyyy+ty(t-1)^{ty^2}y^{ty^2}=t^yy^{y+ty}(t−1)ty2yty2=tyyy+ty
⇒(t−1)ty2yty2−y−ty=ty\Rightarrow (t-1)^{ty^2}y^{ty^2-y-ty}=t^y⇒(t−1)ty2yty2−y−ty=ty.
若 t≥3t\geq 3t≥3 ,则
(t−1)ty2=[(t−1)y]ty>tty>ty(t-1)^{ty^2}=[(t-1)^y]^{ty}>t^{ty}>t^y(t−1)ty2=[(t−1)y]ty>tty>ty.
而 yty2−y−ty≥1y^{ty^2-y-ty}\geq 1yty2−y−ty≥1 ,矛盾.
故 t=2⇒y2y2−3y=2yt=2 \Rightarrow y^{2y^2-3y}=2^yt=2⇒y2y2−3y=2y.
⇒y2y−3=2⇒y=2,x=4\Rightarrow y^{2y-3}=2 \Rightarrow y=2,x=4⇒y2y−3=2⇒y=2,x=4.
综上,(x,y)=(4,2)(x,y)=(4,2)(x,y)=(4,2).xq