Find all positive integers (x,y) such that y2xx+y is a prime number
Problem
Source: Argentina TST Iberoamerican 2009 Problem 4
Tags: number theory unsolved, number theory
26.08.2009 03:09
p(x+y)=y2x.it follows that p|y or p|x. Suppose that both of them are divisable by p. so there exist a,b such that pa||y,pb||x. there are 2 cases: 1-a≠b then from the above equation we'll have 1+min which is absurd. 2- a = b then we'll have (\frac {x}{p^{a}} + \frac {y}{p^{a}}) = p^{a - 1}(\frac {y^2}{p^{2a}})(\frac {x}{p^{a}}) so if \frac {y}{p^{a}} = m,\frac {x}{p^{a}} = n we'll have m^2n\leq{m + n}.we can easily find all the solutions. so assume that p divides only one of them. u can see that in this case we must have p|x. if px' = x u can see that y |x' it follows yt = x'. then we'll have: pt + 1 = y^2t the rest follows. i wish i didn't make any mistake I MUST GO TO BED!!
26.08.2009 22:26
p=\frac {y^2x}{x + y} \iff px+py=y^2x Taking mod x then x|py and taking mod y then y|px (1) py=ax (2) px=by Taking mod p then p|y^2x then either p|x or p|y If p|x then x=kp then in (1) py=akp then y=ak and in (2) pkp=bak then p^2=ab then three cases: (a,b)=(1,p^2) or (p^2,1) or (p,p) If p|y then y=kp then in (2) px=bkp then x=bk and in (1) pkp=abk then p^2=ab then same three cases. Case 1) py=x p=\frac{y^3p}{y(p+1)} \iff p+1=y^2 \iff p=(y+1)(y-1) \Rightarrow y=2 then \boxed{(x,y,p)=(6,2,3)} Case 2) y=px p=\frac{p^2x^3}{x(p+1)} \iff p(p+1)=x^2 \Rightarrow p|x \Rightarrow x=cp \Rightarrow p+1=pc^2 \Rightarrow 1=p(c+1)(c-1) no solution Case 3) y=x then p=\frac{x^3}{2x}=\frac{x^2}{2} then \boxed{(x,y,p)=(2,2,2)}
19.04.2023 22:38
lambruscokid wrote: Find all positive integers (x,y) such that \frac{y^2x}{x+y} is a prime number (x,y)=d x=ad , y=bd \frac{d^{2}×b^{2}×a}{a+b} it must be prime (a,a+b)=1 and (b,a+b)=1 then \frac{d^2}{a+b}=p it must be integer.p×a×b^{2} is prime then b=1 \frac{d^2}{a+1}=p then p×a is prime a=1 or a is prime and p=1 p=1 then d^2=a+1 a=d^2-1 and a is prime then a=3 d=2 \boxed{(x,y,q)=(2,2,2),(6,2,3)}